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Breaking x, y, z into $\frac x2, \frac x2, \frac y3, \frac y3, \frac z2, \frac z2$ and then applying AM GM inequality we get answer to be 432 which is correct.

But when i instead take $2x, x, y, y, y, 2z, z$ (there sum being 7.3=21); i get $3^7\over 4$ to be the max value. Why does the former method work and this doesn't?

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    Remember that just because you show that something is an upper bound, it need not always be the maximum = least upper bound. You have to show that it can be achieved (like by checking the equality case). So you have shown that 3^7/4 is an upper bound, and it is not the least upper bound. – Calvin Lin Aug 25 '24 at 14:21

2 Answers2

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But when i instead take $2x, x, y, y, y, 2z, z$ (their sum being $7 \cdot 3=21$); i get $3^7\over 4$ to be the max value. Why does the former method work and this doesn't?

That's a valid proof that $x^2 y^3 z^2 \leq \frac{3^7}{4}$, yes. But it doesn't prove that some values $(x,y,z)$ actually exist with $x^2 y^3 z^2 = \frac{3^7}{4}$.

Remember that the arithmetic mean and geometric mean of positive numbers are equal only when all those numbers are the same.

In the first method with $\frac x2 + \frac x2 + \frac y3 + \frac y3 + \frac y3 + \frac z2 + \frac z2 = 7$, we know both that $x^2 y^3 z^2 \leq 2^4 3^3 = 432$ and that $x^2 y^3 z^2 = 432$ only if $\frac x2 = \frac y3 = \frac z2$. With the first equation this gives $x=2, y=3, z=2$. Checking, it's true that $2^2 3^3 2^2 = 432$.

In the second approach, $x^2 y^3 z^2 = \frac{3^7}{4}$ only if $2x=x=y=2z=z$. This is clearly impossible for positive $x$ and $z$. So in fact we just get that $x^2 y^3 z^2 < \frac{3^7}{4}$, and this approach doesn't lead to the exact answer.

aschepler
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It is well known equality in AM-GM holds iff all the variables are equal. The proof is here: Proofs of AM-GM inequality if you want to see why. The first way works because you have

$x+y+z=7$

$\Longleftrightarrow 2(\frac{x}{2})+3(\frac{y}{3})+2(\frac{z}{2})=7$

So using AM-GM we have

$\frac{\frac{x}{2}+\frac{x}{2}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{z}{2}+\frac{z}{2}}{7}\ge \sqrt[7]{\frac{x^2y^3z^2}{432}}$

$\Longrightarrow \bigg(\frac{\frac{x}{2}+\frac{x}{2}+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}+\frac{z}{2}+\frac{z}{2}}{7}\bigg)^7\ge \frac{x^2y^3z^2}{432}$

$\Longleftrightarrow \frac{x^2y^3z^2}{432}\le 1$

$\Longleftrightarrow x^2y^3z^2\le 432$

so the maximum value is achieved when

$\frac{x}{2}=\frac{x}{2}=\frac{y}{3}=\frac{y}{3}=\frac{y}{3}=\frac{z}{2}=\frac{z}{2}$, or equivalently $2x=3y=2z$, so plugging these back in the original equation we get $x=z=2$ and $y=3$.

The second approach cannot work because equality cannot hold; proceed as you said, take $2x,x,y,y,y,2z,z$, so we get $$3(x+y+z)=2x+x+y+y+y+2z+z=21$$

so using AM-GM we get

$\frac{2x+x+y+y+y+2z+z}{7}\ge \sqrt[7]{4x^2y^3z^2}$

$\Longleftrightarrow 4x^2y^3z^2\le \big(\frac{21}{7}\big)^7=3^7$

$\Longleftrightarrow x^2y^3z^2\le \frac{3^7}{4}$

with equality iff $2x=x=y=y=y=2z=z$, in particular, $2x=x\Rightarrow x=0$, which is absurd, because $x,y,z>0$. Hope this answers the question :)

Peter
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