<s>[INST] If $x+2y+3z=10$ find the maximum value of $x^2y^2z^6$. If $x+2y+3z=10$ where x,y,z are postive real numbers, find the maximum value of $x^2y^2z^6$.
I started by trying to find some integer solutions. The first thing I found was $(x,y,z)=(1,3,1)$. But the value of $x^2y^2z^6$ was way too small (it was a multiple choice question in the paper I was solving).
The next thing I tried was setting $x=y=z$ because I thought the product of numbers would be highest if they were equal. And I got $x=y=z=\frac{5}{3}$. And so I got $x^2y^2z^6 = (\frac{5}{3})^{10}$
But this answer seems very weird and even my logic is not right, I merely went by intuition.
Please help me solve this.
 [/INST] A.M. $\geq$ G.M.
$$10=x+2y+3z=\frac{x}{2}+\frac{x}{2}+y+y+\frac{z}{2}+\frac{z}{2}+\frac{z}{2}+\frac{z}{2}+\frac{z}{2}+\frac{z}{2} \geq 10{(\frac{x^2}{4}y^2\frac{z^6}{64})^\frac{1}{10}} $$
$$1\geq (\frac{x^2}{4}y^2\frac{z^6}{64})^\frac{1}{10}$$
$$x^2y^2z^6\leq 2^{10}$$
Equilaty holds when $\frac{x}{2}=y=\frac{z}{2}$
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