For $a,b,c>0$ prove that $$\frac{a}{a+\sqrt{(a+b)(a+c)}} + \frac{b}{b+\sqrt{(b+a)(b+c)}} +\frac{c}{c+\sqrt{(c+b)(c+a)}} \leq 1$$
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2026-03-30 15:45:03.1774885503
For $a,b,c>0$ prove that $\frac{a}{a+\sqrt{(a+b)(a+c)}} + \frac{b}{b+\sqrt{(b+a)(b+c)}}+ \frac{c}{c+\sqrt{(c+b)(c+a)}} \leq 1$
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Because by C-S $$\sum_{cyc}\frac{a}{a+\sqrt{(a+b)(a+c)}}\leq\sum_{cyc}\frac{a}{a+\sqrt{ab}+\sqrt{ac}}=\sum_{cyc}\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1$$