<s>[INST] Proving that $\frac{1}{a^2}+\frac{1}{b^2} \geq \frac{8}{(a+b)^2}$ for $a,b>0$ I found something that I'm not quite sure about when trying to prove this inequality.
I've proven that 
$$\dfrac{1}{a}+\dfrac{1}{b}\geq \dfrac{4}{a+b}$$
already. My idea now is to replace $a$ with $a^2$ and $b^2$, so we now have 
$$\dfrac{1}{a^2}+\dfrac{1}{b^2}\geq \dfrac{4}{a^2+b^2}$$
So to prove the required result, we just need to show that 
$$\dfrac{4}{a^2+b^2} \geq \dfrac{8}{(a+b)^2}$$
or equivalently
$$(a+b)^2 \geq 2(a^2+b^2)$$
But this inequality cannot be true since the pair $(a,b)=(1,2)$ doesn't work. If anything, the reverse is always true!
What have I done wrong here?
 [/INST] For $a,b>0$, let
\begin{equation}
A(a,b)=\frac{a+b}{2}, \quad G(a,b)=\sqrt{ab}\,, \quad H(a,b)=\frac{2}{\dfrac1{a}+\dfrac1{b}}
\end{equation}
are respectively the arithmetic, geometric, and harmonic means of two positive numbers $a,b$.
It is commom knowledge that
\begin{equation}
A(a,b)\ge G(a,b) \quad\text{and}\quad H(a,b)=\frac{G^2(a,b)}{A(a,b)}.
\end{equation}
Therefore, we derive
\begin{equation}
A\bigl(a^2,b^2\bigr)\ge G\bigl(a^2,b^2\bigr)=G^2(a,b)
\Longleftrightarrow 1\ge\frac{G^2(a,b)}{A\bigl(a^2,b^2\bigr)}
\end{equation}
and
\begin{equation}
A^2(a,b)\ge G^2(a,b).
\end{equation}
Consequently,  we deduce
\begin{equation}
A^2(a,b)\ge G^2(a,b)\frac{G^2(a,b)}{A\bigl(a^2,b^2\bigr)}
=\frac{G^2\bigl(a^2,b^2\bigr)}{A\bigl(a^2,b^2\bigr)}
=H\bigl(a^2,b^2\bigr)
\end{equation}
which can be reformulated as
\begin{equation}
\biggl(\frac{a+b}{2}\biggr)^2\ge\frac{2}{\dfrac1{a^2}+\dfrac1{b^2}}
\Longleftrightarrow
\dfrac1{a^2}+\dfrac1{b^2}\ge\frac{8}{(a+b)^2}.
\end{equation}
</s>