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?
 A: The problem is simple enough that you can just be mechanical about it
$$
\frac{1}{a^2}+\frac{1}{b^2}\geq\frac{8}{(a+b)^2}\iff(a^2+b^2)(a+b)^2-8a^2b^2\geq0.
$$
The claim now follows because
$$
(a^2+b^2)(a+b)^2-8a^2b^2=(a-b)^2(a^2+4ab+b^2)\geq0,\quad a,b>0.
$$
A: $\dfrac{1}{a^2}+\dfrac{1}{b^2} \ge \dfrac{2}{ab} \ge \dfrac{8}{(a+b)^2} \iff (a+b)^2 \ge 4ab$
A: Use Holder
$$\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)(a+b)(a+b)\ge (1+1)^3=8$$
A: Others have already proved the inequality.
Here, I will answer the following question.

What have I done wrong here?

First, there is nothing wrong in the following part.

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
$$(a+b)^2 \geq 2(a^2+b^2)$$

Here, note that what we can say is that if
$$(a+b)^2 \geq 2(a^2+b^2)$$
is true for $a\gt 0,b\gt 0$, then we know that the inequality is true.
By the way, as you wrote,

But this inequality cannot be true since the pair $(a,b)=(1,2)$ doesn't work.

Now, all we can say is that
$$(a+b)^2 \geq 2(a^2+b^2)$$
is not true.
Hence, we can say nothing about the inequality. (You need to seek another way as others showed.)
It seems that you are misunderstanding something because you are saying

If anything, the reverse is always true!

A: It's a particular case of the power means inequality: the $p$-th power mean of $n$ positive real numbers $a_1,\dots, a_n$ is:
$$M_p(a_1,\dots,a_n)=\biggl(\frac{a_1^p+\dots+a_n^p}n\biggr)^{\!\tfrac1p}.$$
It can be proved that if $p<q$, then $M_p(a_1,\dots,a_n)\le M_q(a_1,\dots,a_n)$, and we have equality if and only if $a_1=\dots=a_n$.
Now the arithmetic mean of $a$ and $b$ is $M_1(a,b)$, and by the above we have the inequality: $\,M_{-2}(a,b)\le  M_1(a,b)$, which is equivalent to $\,M_{-2}(a,b)^2\le  M_1(a,b)^2$, or
$$\frac1{M_{-2}(a,b)^2}=\frac12\biggl(\frac1{a^2}+\frac1{b^2}\biggr)\ge\frac1{M_1(a,b)^2}=\biggl(\frac 2{a+b}\bigg)^2.$$
A: 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}
