Find the minimum of $(1+a^2)(1+b^2)(1+c^2)$ where $a,b,cgeq 0$












3












$begingroup$



Find the minimum of:



$$(1+a^2)(1+b^2)(1+c^2) a,b,cgeq 0$$




Knowing that $$ab+bc+ac=27$$



I tried my best using QM-AM-GM inequalities, Cauchy-Schwarz, etc. I tried also to do it with partial derivatives but it's too long. Thank you for your time :)






inequality contest-math







share|cite|improve this question











$endgroup$












  • $begingroup$
    My initial thought: Let $x=ab,y=bc,z=ac$, so that the constraint function is $x+y+z=27$. Then $xy=ab^2 c=b^2 zimplies b^2=x y/z$ and similarly $a^2=xz/y$, $c^2=yz/x$. That makes the objective function more annoying, but it still seems tractable.
    $endgroup$
    – Semiclassical
    yesterday












  • $begingroup$
    Can you provide a link to the contest? Or some other evidence that the contest is not currently active. I have no reason to think that it is, but we have this rule :-)
    $endgroup$
    – Jyrki Lahtonen
    yesterday










  • $begingroup$
    @JyrkiLahtonen it's just a simulation of math team national italian olympiad. It's not even an official competition :) .The true Olympiads are in May as you can see here olimpiadi.dm.unibo.it (but the site is italian).
    $endgroup$
    – Eureka
    yesterday










  • $begingroup$
    Thanks, Eureka.
    $endgroup$
    – Jyrki Lahtonen
    yesterday
















3












$begingroup$



Find the minimum of:



$$(1+a^2)(1+b^2)(1+c^2) a,b,cgeq 0$$




Knowing that $$ab+bc+ac=27$$



I tried my best using QM-AM-GM inequalities, Cauchy-Schwarz, etc. I tried also to do it with partial derivatives but it's too long. Thank you for your time :)






inequality contest-math







share|cite|improve this question











$endgroup$












  • $begingroup$
    My initial thought: Let $x=ab,y=bc,z=ac$, so that the constraint function is $x+y+z=27$. Then $xy=ab^2 c=b^2 zimplies b^2=x y/z$ and similarly $a^2=xz/y$, $c^2=yz/x$. That makes the objective function more annoying, but it still seems tractable.
    $endgroup$
    – Semiclassical
    yesterday












  • $begingroup$
    Can you provide a link to the contest? Or some other evidence that the contest is not currently active. I have no reason to think that it is, but we have this rule :-)
    $endgroup$
    – Jyrki Lahtonen
    yesterday










  • $begingroup$
    @JyrkiLahtonen it's just a simulation of math team national italian olympiad. It's not even an official competition :) .The true Olympiads are in May as you can see here olimpiadi.dm.unibo.it (but the site is italian).
    $endgroup$
    – Eureka
    yesterday










  • $begingroup$
    Thanks, Eureka.
    $endgroup$
    – Jyrki Lahtonen
    yesterday














3












3








3





$begingroup$



Find the minimum of:



$$(1+a^2)(1+b^2)(1+c^2) a,b,cgeq 0$$




Knowing that $$ab+bc+ac=27$$



I tried my best using QM-AM-GM inequalities, Cauchy-Schwarz, etc. I tried also to do it with partial derivatives but it's too long. Thank you for your time :)






inequality contest-math







share|cite|improve this question











$endgroup$





Find the minimum of:



$$(1+a^2)(1+b^2)(1+c^2) a,b,cgeq 0$$




Knowing that $$ab+bc+ac=27$$



I tried my best using QM-AM-GM inequalities, Cauchy-Schwarz, etc. I tried also to do it with partial derivatives but it's too long. Thank you for your time :)






inequality contest-math




inequality contest-math






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









YuiTo Cheng

2,3144937




2,3144937










asked yesterday









EurekaEureka

712113




712113












  • $begingroup$
    My initial thought: Let $x=ab,y=bc,z=ac$, so that the constraint function is $x+y+z=27$. Then $xy=ab^2 c=b^2 zimplies b^2=x y/z$ and similarly $a^2=xz/y$, $c^2=yz/x$. That makes the objective function more annoying, but it still seems tractable.
    $endgroup$
    – Semiclassical
    yesterday












  • $begingroup$
    Can you provide a link to the contest? Or some other evidence that the contest is not currently active. I have no reason to think that it is, but we have this rule :-)
    $endgroup$
    – Jyrki Lahtonen
    yesterday










  • $begingroup$
    @JyrkiLahtonen it's just a simulation of math team national italian olympiad. It's not even an official competition :) .The true Olympiads are in May as you can see here olimpiadi.dm.unibo.it (but the site is italian).
    $endgroup$
    – Eureka
    yesterday










  • $begingroup$
    Thanks, Eureka.
    $endgroup$
    – Jyrki Lahtonen
    yesterday


















  • $begingroup$
    My initial thought: Let $x=ab,y=bc,z=ac$, so that the constraint function is $x+y+z=27$. Then $xy=ab^2 c=b^2 zimplies b^2=x y/z$ and similarly $a^2=xz/y$, $c^2=yz/x$. That makes the objective function more annoying, but it still seems tractable.
    $endgroup$
    – Semiclassical
    yesterday












  • $begingroup$
    Can you provide a link to the contest? Or some other evidence that the contest is not currently active. I have no reason to think that it is, but we have this rule :-)
    $endgroup$
    – Jyrki Lahtonen
    yesterday










  • $begingroup$
    @JyrkiLahtonen it's just a simulation of math team national italian olympiad. It's not even an official competition :) .The true Olympiads are in May as you can see here olimpiadi.dm.unibo.it (but the site is italian).
    $endgroup$
    – Eureka
    yesterday










  • $begingroup$
    Thanks, Eureka.
    $endgroup$
    – Jyrki Lahtonen
    yesterday
















$begingroup$
My initial thought: Let $x=ab,y=bc,z=ac$, so that the constraint function is $x+y+z=27$. Then $xy=ab^2 c=b^2 zimplies b^2=x y/z$ and similarly $a^2=xz/y$, $c^2=yz/x$. That makes the objective function more annoying, but it still seems tractable.
$endgroup$
– Semiclassical
yesterday






$begingroup$
My initial thought: Let $x=ab,y=bc,z=ac$, so that the constraint function is $x+y+z=27$. Then $xy=ab^2 c=b^2 zimplies b^2=x y/z$ and similarly $a^2=xz/y$, $c^2=yz/x$. That makes the objective function more annoying, but it still seems tractable.
$endgroup$
– Semiclassical
yesterday














$begingroup$
Can you provide a link to the contest? Or some other evidence that the contest is not currently active. I have no reason to think that it is, but we have this rule :-)
$endgroup$
– Jyrki Lahtonen
yesterday




$begingroup$
Can you provide a link to the contest? Or some other evidence that the contest is not currently active. I have no reason to think that it is, but we have this rule :-)
$endgroup$
– Jyrki Lahtonen
yesterday












$begingroup$
@JyrkiLahtonen it's just a simulation of math team national italian olympiad. It's not even an official competition :) .The true Olympiads are in May as you can see here olimpiadi.dm.unibo.it (but the site is italian).
$endgroup$
– Eureka
yesterday




$begingroup$
@JyrkiLahtonen it's just a simulation of math team national italian olympiad. It's not even an official competition :) .The true Olympiads are in May as you can see here olimpiadi.dm.unibo.it (but the site is italian).
$endgroup$
– Eureka
yesterday












$begingroup$
Thanks, Eureka.
$endgroup$
– Jyrki Lahtonen
yesterday




$begingroup$
Thanks, Eureka.
$endgroup$
– Jyrki Lahtonen
yesterday










2 Answers
2






active

oldest

votes


















13












$begingroup$

We note that $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$, and $c$. We rewrite it by using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$:
$$begin{align}
(1+a^2)(1+b^2)(1+c^2)&=(ab+bc+ca-1)^2+(a+b+c-abc)^2\
&=26^2+(a+b+c-abc)^2geq 26^2=676
end{align}$$
where the condition $ab+bc+ac=27$ has been applied.



Hence the minimum is $676$ as soon as we show that there are $a,b,cgeq 0$ such that $ab+bc+ac=27$ and $a+b+c=abc$.
Can you take it from here?






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    @Robert Z Amazing! +1
    $endgroup$
    – Michael Rozenberg
    yesterday






  • 1




    $begingroup$
    How had you this idea ? +1
    $endgroup$
    – Jean Marie
    yesterday






  • 1




    $begingroup$
    $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$ and $c$. Try to rewrite it using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$.
    $endgroup$
    – Robert Z
    yesterday






  • 4




    $begingroup$
    I have a rather direct proof of the identity used by @Robert Z : use twice what is called Diophantes equality $(a^2+b^2)(c^2+d^2)=(ac−bd)^2+(ad+bc)^2$ (which can be explained as two ways of writing the squared norm ( = squared module) of $(a+ib)(c+id)$)... Or, said in an even simpler way take the squared norm of both sides of relationship $(1+ia)(1+ib)(1+ic)=(1-ab-bc-ca)+(a+b+c-abc)i$.
    $endgroup$
    – Jean Marie
    yesterday








  • 1




    $begingroup$
    Good job. You even managed to convince Michael about the usefulness of elementary symmetric polynomials - something I failed the last time I tried.
    $endgroup$
    – Jyrki Lahtonen
    yesterday



















3












$begingroup$

A simple supplement to Robert's nice answer. With the aid of the following plot I'm sure you all can prove that when $a+b+c=12=abc$ the polynomial
$$(x-a)(x-b)(x-c)=x^3-12x^2+27x-12$$
has three positive zeros $a,b,c$ :-)



enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Adding this because it was not immediately obvious to me how to prove that Robert's lower bound can be achieved. I'm sure there are nice trick ways of doing that. But, it really is simple.
    $endgroup$
    – Jyrki Lahtonen
    yesterday










  • $begingroup$
    Thanks for this supplement (+1)
    $endgroup$
    – Robert Z
    16 hours ago












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









13












$begingroup$

We note that $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$, and $c$. We rewrite it by using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$:
$$begin{align}
(1+a^2)(1+b^2)(1+c^2)&=(ab+bc+ca-1)^2+(a+b+c-abc)^2\
&=26^2+(a+b+c-abc)^2geq 26^2=676
end{align}$$
where the condition $ab+bc+ac=27$ has been applied.



Hence the minimum is $676$ as soon as we show that there are $a,b,cgeq 0$ such that $ab+bc+ac=27$ and $a+b+c=abc$.
Can you take it from here?






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    @Robert Z Amazing! +1
    $endgroup$
    – Michael Rozenberg
    yesterday






  • 1




    $begingroup$
    How had you this idea ? +1
    $endgroup$
    – Jean Marie
    yesterday






  • 1




    $begingroup$
    $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$ and $c$. Try to rewrite it using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$.
    $endgroup$
    – Robert Z
    yesterday






  • 4




    $begingroup$
    I have a rather direct proof of the identity used by @Robert Z : use twice what is called Diophantes equality $(a^2+b^2)(c^2+d^2)=(ac−bd)^2+(ad+bc)^2$ (which can be explained as two ways of writing the squared norm ( = squared module) of $(a+ib)(c+id)$)... Or, said in an even simpler way take the squared norm of both sides of relationship $(1+ia)(1+ib)(1+ic)=(1-ab-bc-ca)+(a+b+c-abc)i$.
    $endgroup$
    – Jean Marie
    yesterday








  • 1




    $begingroup$
    Good job. You even managed to convince Michael about the usefulness of elementary symmetric polynomials - something I failed the last time I tried.
    $endgroup$
    – Jyrki Lahtonen
    yesterday
















13












$begingroup$

We note that $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$, and $c$. We rewrite it by using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$:
$$begin{align}
(1+a^2)(1+b^2)(1+c^2)&=(ab+bc+ca-1)^2+(a+b+c-abc)^2\
&=26^2+(a+b+c-abc)^2geq 26^2=676
end{align}$$
where the condition $ab+bc+ac=27$ has been applied.



Hence the minimum is $676$ as soon as we show that there are $a,b,cgeq 0$ such that $ab+bc+ac=27$ and $a+b+c=abc$.
Can you take it from here?






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    @Robert Z Amazing! +1
    $endgroup$
    – Michael Rozenberg
    yesterday






  • 1




    $begingroup$
    How had you this idea ? +1
    $endgroup$
    – Jean Marie
    yesterday






  • 1




    $begingroup$
    $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$ and $c$. Try to rewrite it using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$.
    $endgroup$
    – Robert Z
    yesterday






  • 4




    $begingroup$
    I have a rather direct proof of the identity used by @Robert Z : use twice what is called Diophantes equality $(a^2+b^2)(c^2+d^2)=(ac−bd)^2+(ad+bc)^2$ (which can be explained as two ways of writing the squared norm ( = squared module) of $(a+ib)(c+id)$)... Or, said in an even simpler way take the squared norm of both sides of relationship $(1+ia)(1+ib)(1+ic)=(1-ab-bc-ca)+(a+b+c-abc)i$.
    $endgroup$
    – Jean Marie
    yesterday








  • 1




    $begingroup$
    Good job. You even managed to convince Michael about the usefulness of elementary symmetric polynomials - something I failed the last time I tried.
    $endgroup$
    – Jyrki Lahtonen
    yesterday














13












13








13





$begingroup$

We note that $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$, and $c$. We rewrite it by using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$:
$$begin{align}
(1+a^2)(1+b^2)(1+c^2)&=(ab+bc+ca-1)^2+(a+b+c-abc)^2\
&=26^2+(a+b+c-abc)^2geq 26^2=676
end{align}$$
where the condition $ab+bc+ac=27$ has been applied.



Hence the minimum is $676$ as soon as we show that there are $a,b,cgeq 0$ such that $ab+bc+ac=27$ and $a+b+c=abc$.
Can you take it from here?






share|cite|improve this answer











$endgroup$



We note that $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$, and $c$. We rewrite it by using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$:
$$begin{align}
(1+a^2)(1+b^2)(1+c^2)&=(ab+bc+ca-1)^2+(a+b+c-abc)^2\
&=26^2+(a+b+c-abc)^2geq 26^2=676
end{align}$$
where the condition $ab+bc+ac=27$ has been applied.



Hence the minimum is $676$ as soon as we show that there are $a,b,cgeq 0$ such that $ab+bc+ac=27$ and $a+b+c=abc$.
Can you take it from here?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









Robert ZRobert Z

101k1072145




101k1072145








  • 2




    $begingroup$
    @Robert Z Amazing! +1
    $endgroup$
    – Michael Rozenberg
    yesterday






  • 1




    $begingroup$
    How had you this idea ? +1
    $endgroup$
    – Jean Marie
    yesterday






  • 1




    $begingroup$
    $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$ and $c$. Try to rewrite it using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$.
    $endgroup$
    – Robert Z
    yesterday






  • 4




    $begingroup$
    I have a rather direct proof of the identity used by @Robert Z : use twice what is called Diophantes equality $(a^2+b^2)(c^2+d^2)=(ac−bd)^2+(ad+bc)^2$ (which can be explained as two ways of writing the squared norm ( = squared module) of $(a+ib)(c+id)$)... Or, said in an even simpler way take the squared norm of both sides of relationship $(1+ia)(1+ib)(1+ic)=(1-ab-bc-ca)+(a+b+c-abc)i$.
    $endgroup$
    – Jean Marie
    yesterday








  • 1




    $begingroup$
    Good job. You even managed to convince Michael about the usefulness of elementary symmetric polynomials - something I failed the last time I tried.
    $endgroup$
    – Jyrki Lahtonen
    yesterday














  • 2




    $begingroup$
    @Robert Z Amazing! +1
    $endgroup$
    – Michael Rozenberg
    yesterday






  • 1




    $begingroup$
    How had you this idea ? +1
    $endgroup$
    – Jean Marie
    yesterday






  • 1




    $begingroup$
    $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$ and $c$. Try to rewrite it using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$.
    $endgroup$
    – Robert Z
    yesterday






  • 4




    $begingroup$
    I have a rather direct proof of the identity used by @Robert Z : use twice what is called Diophantes equality $(a^2+b^2)(c^2+d^2)=(ac−bd)^2+(ad+bc)^2$ (which can be explained as two ways of writing the squared norm ( = squared module) of $(a+ib)(c+id)$)... Or, said in an even simpler way take the squared norm of both sides of relationship $(1+ia)(1+ib)(1+ic)=(1-ab-bc-ca)+(a+b+c-abc)i$.
    $endgroup$
    – Jean Marie
    yesterday








  • 1




    $begingroup$
    Good job. You even managed to convince Michael about the usefulness of elementary symmetric polynomials - something I failed the last time I tried.
    $endgroup$
    – Jyrki Lahtonen
    yesterday








2




2




$begingroup$
@Robert Z Amazing! +1
$endgroup$
– Michael Rozenberg
yesterday




$begingroup$
@Robert Z Amazing! +1
$endgroup$
– Michael Rozenberg
yesterday




1




1




$begingroup$
How had you this idea ? +1
$endgroup$
– Jean Marie
yesterday




$begingroup$
How had you this idea ? +1
$endgroup$
– Jean Marie
yesterday




1




1




$begingroup$
$(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$ and $c$. Try to rewrite it using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$.
$endgroup$
– Robert Z
yesterday




$begingroup$
$(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$ and $c$. Try to rewrite it using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$.
$endgroup$
– Robert Z
yesterday




4




4




$begingroup$
I have a rather direct proof of the identity used by @Robert Z : use twice what is called Diophantes equality $(a^2+b^2)(c^2+d^2)=(ac−bd)^2+(ad+bc)^2$ (which can be explained as two ways of writing the squared norm ( = squared module) of $(a+ib)(c+id)$)... Or, said in an even simpler way take the squared norm of both sides of relationship $(1+ia)(1+ib)(1+ic)=(1-ab-bc-ca)+(a+b+c-abc)i$.
$endgroup$
– Jean Marie
yesterday






$begingroup$
I have a rather direct proof of the identity used by @Robert Z : use twice what is called Diophantes equality $(a^2+b^2)(c^2+d^2)=(ac−bd)^2+(ad+bc)^2$ (which can be explained as two ways of writing the squared norm ( = squared module) of $(a+ib)(c+id)$)... Or, said in an even simpler way take the squared norm of both sides of relationship $(1+ia)(1+ib)(1+ic)=(1-ab-bc-ca)+(a+b+c-abc)i$.
$endgroup$
– Jean Marie
yesterday






1




1




$begingroup$
Good job. You even managed to convince Michael about the usefulness of elementary symmetric polynomials - something I failed the last time I tried.
$endgroup$
– Jyrki Lahtonen
yesterday




$begingroup$
Good job. You even managed to convince Michael about the usefulness of elementary symmetric polynomials - something I failed the last time I tried.
$endgroup$
– Jyrki Lahtonen
yesterday











3












$begingroup$

A simple supplement to Robert's nice answer. With the aid of the following plot I'm sure you all can prove that when $a+b+c=12=abc$ the polynomial
$$(x-a)(x-b)(x-c)=x^3-12x^2+27x-12$$
has three positive zeros $a,b,c$ :-)



enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Adding this because it was not immediately obvious to me how to prove that Robert's lower bound can be achieved. I'm sure there are nice trick ways of doing that. But, it really is simple.
    $endgroup$
    – Jyrki Lahtonen
    yesterday










  • $begingroup$
    Thanks for this supplement (+1)
    $endgroup$
    – Robert Z
    16 hours ago
















3












$begingroup$

A simple supplement to Robert's nice answer. With the aid of the following plot I'm sure you all can prove that when $a+b+c=12=abc$ the polynomial
$$(x-a)(x-b)(x-c)=x^3-12x^2+27x-12$$
has three positive zeros $a,b,c$ :-)



enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Adding this because it was not immediately obvious to me how to prove that Robert's lower bound can be achieved. I'm sure there are nice trick ways of doing that. But, it really is simple.
    $endgroup$
    – Jyrki Lahtonen
    yesterday










  • $begingroup$
    Thanks for this supplement (+1)
    $endgroup$
    – Robert Z
    16 hours ago














3












3








3





$begingroup$

A simple supplement to Robert's nice answer. With the aid of the following plot I'm sure you all can prove that when $a+b+c=12=abc$ the polynomial
$$(x-a)(x-b)(x-c)=x^3-12x^2+27x-12$$
has three positive zeros $a,b,c$ :-)



enter image description here






share|cite|improve this answer











$endgroup$



A simple supplement to Robert's nice answer. With the aid of the following plot I'm sure you all can prove that when $a+b+c=12=abc$ the polynomial
$$(x-a)(x-b)(x-c)=x^3-12x^2+27x-12$$
has three positive zeros $a,b,c$ :-)



enter image description here







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








answered yesterday


























community wiki





Jyrki Lahtonen













  • $begingroup$
    Adding this because it was not immediately obvious to me how to prove that Robert's lower bound can be achieved. I'm sure there are nice trick ways of doing that. But, it really is simple.
    $endgroup$
    – Jyrki Lahtonen
    yesterday










  • $begingroup$
    Thanks for this supplement (+1)
    $endgroup$
    – Robert Z
    16 hours ago


















  • $begingroup$
    Adding this because it was not immediately obvious to me how to prove that Robert's lower bound can be achieved. I'm sure there are nice trick ways of doing that. But, it really is simple.
    $endgroup$
    – Jyrki Lahtonen
    yesterday










  • $begingroup$
    Thanks for this supplement (+1)
    $endgroup$
    – Robert Z
    16 hours ago
















$begingroup$
Adding this because it was not immediately obvious to me how to prove that Robert's lower bound can be achieved. I'm sure there are nice trick ways of doing that. But, it really is simple.
$endgroup$
– Jyrki Lahtonen
yesterday




$begingroup$
Adding this because it was not immediately obvious to me how to prove that Robert's lower bound can be achieved. I'm sure there are nice trick ways of doing that. But, it really is simple.
$endgroup$
– Jyrki Lahtonen
yesterday












$begingroup$
Thanks for this supplement (+1)
$endgroup$
– Robert Z
16 hours ago




$begingroup$
Thanks for this supplement (+1)
$endgroup$
– Robert Z
16 hours ago


















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