Prove that: $\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}+\sqrt{3(a^2+b^2+c^2-ab-bc-ca)}\ge \frac{3}{2}$ Problem:

Given non- negative real numbers such that: $ab+bc+ca>0: a+b+c=3.$ Prove that: $$\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}+\sqrt{3(a^2+b^2+c^2-ab-bc-ca)}\ge \frac{3}{2}$$
My approach:

We have: $a^2+b^2+c^2\ge ab+bc+ca$. It is desired to show that: $$\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}\ge\frac{3}{2}$$.
But it seems not work by calculation example. I hope we can get nice proof for problem. Thanks!

The following stronger inequality is also true.

Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c=3$ and $ab+ac+bc\neq0$. Prove that:
$$\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}+\frac{7}{27}\sqrt{a^2+b^2+c^2-ab-bc-ca}\ge \frac{3}{2}.$$

Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, the condition and $\sum\limits_{cyc}(a^2-ab)=9u^2-9v^2$ don't depend on $w^3$ and $f(w^3)=\sum\limits_{cyc}\frac{1}{a^2+b^2}$ increases, which by $uvw$ says that it's enough to prove our inequality in two cases:

*

*$w^3=0$;


*Two variables are equal, which easy to check.
Can you take it from here?
About $uvw$ see here:
https://artofproblemsolving.com/community/c6h278791
