Find the maximum constant such that the inequality

Question

Let $a;b>0$. Find the maximum constant such that the inequality $$\frac{1}{a^2+b^2}+\frac{1}{a^2}+\frac{1}{b^2}\ge \frac{8+2k}{\left(a+b\right)^2}$$

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Let $a=1$ then we have: $-\frac{k-1}{2a^2}\ge 0\Leftrightarrow k\le 1$. So we will prove $k=1$ is the maximum constant.

$$\frac{\left(a-b\right)^2\left(a^4+4a^3b+a^2b^2+4ab^3+b^4\right)}{a^2b^2\left(a+b\right)^2\left(a^2+b^2\right)}\ge 0$$

Is that true ?

Answer 1

Your solution is right.

The proof of the final inequality we can make also by AM-GM.

Indeed, $$\frac{1}{a^2+b^2}+\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{a^2+b^2}+\frac{a^2+b^2}{a^2b^2}=$$ $$=\frac{1}{a^2+b^2}+4\cdot\frac{a^2+b^2}{4a^2b^2}\geq5\sqrt[5]{\frac{1}{a^2+b^2}\left(\frac{a^2+b^2}{4a^2b^2}\right)^4}=$$ $$=5\sqrt[5]{\frac{(a^2+b^2)^3}{256a^8b^8}}\geq5\sqrt[5]{\frac{(2ab)^3}{256a^8b^8}}=5\sqrt[5]{\frac{1}{32a^5b^5}}=\frac{10}{4ab}\geq\frac{10}{(a+b)^2}.$$