Quidditch wrote: Let $a, b, c$ be positive real numbers. Prove that $$\frac{a}{a+\sqrt{(a+b)(a+c)}}+\frac{b}{b+\sqrt{(b+c)(b+a)}}+\frac{c}{c+\sqrt{(c+a)(c+b)}}\leq\frac{2a^2+ab}{(b+\sqrt{ca}+c)^2}+\frac{2b^2+bc}{(c+\sqrt{ab}+a)^2}+\frac{2c^2+ca}{(a+\sqrt{bc}+b)^2}.$$ Because $$\sum_{cyc}\frac{a}{a+\sqrt{(a+b)(a+c)}}=\sum_{cyc}\frac{a\left(\sqrt{(a+b)(a+c)}-a\right)}{ab+ac+bc}\leq\sum_{cyc}\frac{a\left(\frac{2a+b+c}{2}-a\right)}{ab+ac+bc}=$$$$=1\leq\sum_{cyc}\frac{2a^2+ab}{(a+b+c)(b+2c)}=\sum_{cyc}\frac{2a^2+ab}{(b+a+c)(b+c+c)}\leq\sum_{cyc}\frac{2a^2+ab}{\left(b+\sqrt{ac}+c\right)^2}.$$

The following is nice: $$\frac{a+\sqrt{bc}}{a+\sqrt{(a+b)(a+c)}}+\frac{b+\sqrt{ca}}{b+\sqrt{(b+c)(b+a)}}+\frac{c+\sqrt{ab}}{c+\sqrt{(c+a)(c+b)}}\leq2$$

For #3, C-S or AM-GM helps well

arqady wrote: Quidditch wrote: Let $a, b, c$ be positive real numbers. Prove that $$\frac{a}{a+\sqrt{(a+b)(a+c)}}+\frac{b}{b+\sqrt{(b+c)(b+a)}}+\frac{c}{c+\sqrt{(c+a)(c+b)}}\leq\frac{2a^2+ab}{(b+\sqrt{ca}+c)^2}+\frac{2b^2+bc}{(c+\sqrt{ab}+a)^2}+\frac{2c^2+ca}{(a+\sqrt{bc}+b)^2}.$$ Because $$\sum_{cyc}\frac{a}{a+\sqrt{(a+b)(a+c)}}=\sum_{cyc}\frac{a\left(\sqrt{(a+b)(a+c)}-a\right)}{ab+ac+bc}\leq\sum_{cyc}\frac{a\left(\frac{2a+b+c}{2}-a\right)}{ab+ac+bc}=$$$$=1\leq\sum_{cyc}\frac{2a^2+ab}{(a+b+c)(b+2c)}=\sum_{cyc}\frac{2a^2+ab}{(b+a+c)(b+c+c)}\leq\sum_{cyc}\frac{2a^2+ab}{\left(b+\sqrt{ac}+c\right)^2}.$$ Art