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:
$ab\le \frac{\left(a+b\right)^2}{2}\:\rightarrow \:2\le \frac{\left(a+b\right)^2}{4}\:\rightarrow \:\frac{1}{2}\le \frac{\left(a+b\right)^2}{16}\:\rightarrow \:\frac{3}{2\left(a+b\right)^2}\le \:\frac{3}{16}$

Substitute the given restriction into the proposed expression.
Then your problem reduces to study a single-variable real-valued function:
\begin{align*}
\frac{3}{2(a + b)^{2}} & = \frac{3}{2\left(a + \dfrac{2}{a}\right)^{2}}\\\\
& = \frac{3a^{2}}{2(a^{2} + 2)^{2}}\\\\
& = \frac{3}{2}\left(\frac{a}{a^{2} + 2}\right)^{2}
\end{align*}
Since the quadratic function is increasing, it suffices to study when the argument attains its max.
More precisely, according to the AM-GM inequality, we can conclude that
\begin{align*}
a^{2} + 2 \geq 2\sqrt{2a^{2}} = 2\sqrt{2}|a| \Longleftrightarrow \left|\frac{a}{a^{2} + 2}\right| \leq \frac{1}{2\sqrt{2}}
\end{align*}
Hence the maximum value attained is given by:
\begin{align*}
\frac{3}{2(a + b)^{2}} \leq \frac{3}{2}\left(\frac{1}{2\sqrt{2}}\right)^{2} = \frac{3}{16}
\end{align*}
In order to determine $a$ which corresponds to the maximum, you can solve the equation:
\begin{align*}
a^{2} - 2\sqrt{2}|a| + 2 = 0
\end{align*}
Once you know $a$, you also know $b$ and you are done.
Hopefully this helps !
