A inequality proposed at Zhautykov Olympiad 2008 Let be $a,b,c >0$ with $abc=1$. Prove that: $$\sum_{cyc}{\frac{1}{(a+b)b}} \geq \frac{3}{2}.$$ $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$. Our inequality becomes: $$\sum_{cyc}{\frac{z^2}{zx+y^2}} \geq \frac{3}{2}.$$ Now we use that: $z^2+x^2 \geq 2zx.$ $$\sum_{cyc}{\frac{z^2}{zx+y^2}} \geq \sum_{cyc}{\frac{2z^2}{z^2+x^2+2y^2}} \geq \frac{3}{2}.$$ Now applying Cauchy Schwarz we obtain the desired result . What I wrote can be found […]

$a, b,c $ are positive real numbers such that $a+b+c = 3$, prove that :$a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6$ Any ideas ?

Let $a,b,c$ are non-negative numbers, such that $a+b+c = 3$. Prove that $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$ Here’s my idea: $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$ $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(ab + bc + ca)$ $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) – 2(ab […]

Given that $xyz=1$, prove that $$\frac{x}{1+x^4}+\frac{y}{1+y^4}+\frac{z}{1+z^4}\le\frac{3}{2}.$$ I proved this with Muirhead’s inequality, but: is there a better/more elegant way?

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