parmenides51 wrote: Let $x, y, z \in R^+$. Prove that $$\frac{x}{x +\sqrt{(x + y)(x + z)}}+\frac{y}{y +\sqrt{(y + z)(y + x)}}+\frac{z}{z +\sqrt{(x + z)(z + y)}} \le 1$$ Use C-S

We use Cauchy in 2 different ways to solve this problem. #1: $\sqrt{(x+y)(x+z)} \geq \sqrt{xy}+\sqrt{xz}$ So $$\sum \frac{x}{x +\sqrt{(x + y)(x + z)}} \leq \sum \frac{x}{x+\sqrt{xy}+\sqrt{xz}}$$$$=\sum \frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1$$#2: $\sqrt{(x+y)(x+z)} \geq x+\sqrt{yz}$ $$\sum \frac{x}{x +\sqrt{(x + y)(x + z)}} \leq \sum \frac{x}{2x+\sqrt{yz}}$$Let $a=\frac{\sqrt{yz}}{x}, b=\frac{\sqrt{zx}}{y}, c=\frac{\sqrt{xy}}{z}$, and therefore $abc=1.$ We thus want to prove $$\sum \frac{1}{a+2} \leq 1$$Which is equivalent to $$ab+bc+ca \geq 3,$$which is obvious by AM-GM.

Let $x, y, z \in R^+$. Prove that $$\frac{x}{2x +\sqrt{(x + y)(x + z)}}+\frac{y}{2y +\sqrt{(y + z)(y + x)}}+\frac{z}{2z +\sqrt{(x + z)(z + y)}} \le \frac{3}{4}$$

parmenides51 wrote: Let $x, y, z \in R^+$. Prove that $$\frac{x}{x +\sqrt{(x + y)(x + z)}}+\frac{y}{y +\sqrt{(y + z)(y + x)}}+\frac{z}{z +\sqrt{(x + z)(z + y)}} \le 1$$ https://artofproblemsolving.com/community/c6h299899p1650486 https://artofproblemsolving.com/community/c6h1523250p9111773