Given $ab+bc+ca=3abc$, prove $\sqrt{\frac{a+b}{c(a^2+b^2 )}}+\sqrt{\frac{b+c}{a(b^2+c^2)}}+\sqrt{\frac{c+a}{b(c^2+a^2 )}}\leq 3$

Question

$a, b, c$ are positive real numbers such that $ab+bc+ca=3abc$

Prove∶
$$\sqrt{\frac{a+b}{c(a^2+b^2 )}}+\sqrt{\frac{b+c}{a(b^2+c^2)}}+\sqrt{\frac{c+a}{b(c^2+a^2 )}}\;\;\leq\; 3$$

Answer 1

We need to prove that $$\sum\limits_{cyc}\sqrt{\frac{a+b}{c(a^2+b^2)}}\leq\frac{ab+ac+bc}{abc}$$ or $$\sum\limits_{cyc}\sqrt{\frac{ab(a+b)}{a^2+b^2}}\leq\frac{ab+ac+bc}{\sqrt{abc}}$$ By C-S $\left(\sum\limits_{cyc}\sqrt{\frac{ab(a+b)}{a^2+b^2}}\right)^2\leq(ab+ac+bc)\sum\limits_{cyc}\frac{a+b}{a^2+b^2}$.

Thus, it remains to prove that $\sum\limits_{cyc}\frac{a+b}{a^2+b^2}\leq\frac{ab+ac+bc}{abc}$ or $\sum\limits_{cyc}\frac{a+b}{a^2+b^2}\leq\frac{1}{2}\sum\limits_{cyc}\left(\frac{1}{a}+\frac{1}{b}\right)$,

which is just $\sum\limits_{cyc}\frac{(a+b)(a-b)^2}{2ab(a^2+b^2)}\geq0$.

Done!