# Leo Giugiuc's Cyclic Inequality in Square Roots

### Problem

### Solution

We know there is acute $\Delta ABC$ such that $\sqrt{x+y}=c,$ $\sqrt{y+z}=a,$ $\sqrt{z+x}=v.$ Then $\displaystyle \sqrt{\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}}=2R,$ where $R$ is the circumradius. By the Law of Sines, $a=2R\sin A,$ $b=2R\sin B,$ $c=2R\sin C.$

The first problem reduces to $\sin A+\sin B+\sin C\gt 2.$

Observe that $\sin(x)$ is strictly concave on $\left[0,\frac{\pi}{2}\right].$ Since the sequence $\left(\frac{\pi}{2},\frac{\pi}{2},0\right)$ strictly majorizes the sequence $(A,B,C),$ by *Karamata's theorem*,

$\sin A+\sin B+\sin C\gt \sin\frac{\pi}{2}+\sin\frac{\pi}{2}+\sin 0=2.$

For the second part, given $k\gt 2,$ we shall prove the existence of acute $\Delta ABC$ for which $\sin A+\sin B+\sin C\lt k.$

For every $\displaystyle x\in\left(\frac{\pi}{4},\frac{\pi}{2}\right)$ consider $\Delta ABC$ with $A=\pi-2x$ and $B=C=x.$ Then

$\displaystyle \lim_{x\to\frac{\pi}{2}^{-}}(\sin (\pi-2x)+2\sin x)=2\lt k.$

### Acknowledgment

Leo Giugiuc has kindly emailed me a problem of his, along with his own solution. The problem was included in the Romania TST 2018 (5).

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