Let $a,b\in \mathbb{R}$ such that $ab=2$. Find the max value of $\frac{3}{2\left(a+b\right)^2}$ and $a, b$ where max is attained, without calculus. My thinking:
By Arithmetic and Geometric mean inequality (AGM):
$ab\le \left(\frac{a+b}{2}\right)^2$
We know $ab=2$
$\rightarrow$ $2\le \left(\frac{a+b}{2}\right)^2$
$\rightarrow$ $2\le \frac{a^2+b^2}{4}$
$\rightarrow$ $\frac{1}{2}\le \:\frac{1}{\frac{a^2+b^2}{4}}$
$\rightarrow$ $\frac{1}{2}\le \:\frac{4}{a^2+b^2}$
$\rightarrow$ $\frac{1}{2}\left(\frac{3}{a^2+b^2}\right)\le \:\frac{4}{a^2+b^2}\left(\frac{3}{a^2+b^2}\right)$
$\rightarrow$ $\frac{3}{2\left(a+b\right)^2}\le \frac{12}{\left(a+b\right)^4}$
Therefore the max value is $\frac{12}{\left(a+b\right)^4}$
Maximum value is attained when $a=b$:
So we can write:
$\rightarrow$ $\frac{3}{2\left(a\right)^2}=\frac{12}{\left(a\right)^4}$
$\rightarrow$ $3a^4=24a^2$
$\rightarrow$ $a=0, -2\sqrt{2},+2\sqrt{2}$
Therefore the max value of $a,b$ are $0,-2\sqrt{2},+2\sqrt{2}$
I'm not completely sure about my answer, if anyone could provide some feedback, that would be amazing! Thanks in advance.
EDIT:
$ab\le \frac{\left(a+b\right)^2}{2}\:\rightarrow \:2\le \frac{\left(a+b\right)^2}{4}\:\rightarrow \:\frac{1}{2}\le \frac{\left(a+b\right)^2}{16}\:\rightarrow \:\frac{3}{2\left(a+b\right)^2}\le \:\frac{3}{16}$

Why you found the maximal value of $a, b$? The question is asking $\frac3{2(a+b)^2}$!
Using AM-GM is, yes, the best approach. Here is my answer.
Since using AM-GM, $8=4ab\le(a+b)^2$, so $\frac1{(a+b)^2}\le\frac18$.
So the maxima of $\frac3{2(a+b)^2}$ is $\frac32\cdot\frac18=\frac3{16}$ when $a=b=\pm\sqrt2$.
