29-th Balkan Mathematical Olympiad

We will obtain the inequality by adding the inequalities $$(x+y)\sqrt{(z+x)(z+y)}\ge 2xy+yz+zx$$ for cyclic permutation of $x$, $y$, $z$. Squaring both sides of this inequality we obtain $$(x+y{)}^2(z+x)(z+y)\ge 4x^2{y}^2+y^2{z}^2+z^2{x}^2+4xy{z}^2+4x^{2}yz+2xy{z}^2$$ which is equivalent to $$x^{3}y+x{y}^3+z(x^3+y^3)\ge 2x^2{y}^2+xyz(x+y)$$ which can be rearranged to $$(xy+yz+zx)(x-y{)}^2\ge 0$$ which is clearly true.

---

Alternative solution.

For positive real numbers $x$, $y$, $z$ there exists a triangle with the side lengths $\sqrt{x+y}$, $\sqrt{y+z}$, $\sqrt{z+x}$ and the area $K=\sqrt{xy+yz+zx}/2$. The existence of the triangle is clear by simple checking of the triangle inequality. To prove the area formula, we have $$K=\frac{1}{2}\sqrt{x+y}\sqrt{z+x}\cos \alpha ,$$ where $\alpha$ is the angle between the sides of length $\sqrt{x+y}$ and $\sqrt{z+x}$. On the other hand, from the law of cosines we have $$\cos \alpha =\frac{x+y+z+x-y-z}{2\sqrt{(x+y)(z+x)}}=\frac{x}{\sqrt{(x+y)(z+x)}},$$ and $$\sin \alpha =\sqrt{1-{\cos }^2\alpha }=\frac{\sqrt{xy+yz+zx}}{\sqrt{(x+y)(x+z)}}.$$ Now the inequality is equivalent to $$\sqrt{x+y}\sqrt{y+z}\sqrt{z+x}\sum _{\text{cyc}}\sqrt{x+y}\ge 16K^2.$$ This can be rewritten as $$\frac{\sqrt{x+y}\sqrt{y+z}\sqrt{z+x}}{4K}\ge 2\frac{K}{{\sum }_{\text{cyc}}\sqrt{x+y}/2}$$ to become the Euler inequality $R\ge 2r$.

---

Alternative solution.

We have $$\sum _{\text{cyc}}(x+y)\sqrt{(z+x)(z+y)}=(x+y)(y+z)(z+x)\sum _{\text{cyc}}\frac{1}{\sqrt{(z+x)(z+y)}}$$ We have on one side $$\sum _{\text{cyc}}\frac{1}{\sqrt{(z+x)(z+y)}}\ge \sum _{\text{cyc}}\frac{2}{x+y+2z}\ge \frac{9}{2}\frac{1}{x+y+z}$$ We know that $$(x+y)(y+z)(z+x)=(xy+yz+zx)(x+y+z)-xyz.$$ On the other side we have $$(xy+yz+zx)(x+y+z)\ge 9\sqrt[3]{x^2{y}^2{z}^2}\sqrt{xyz}=9xyz.$$ Thus $$\sum _{\text{cyc}}(x+y)\sqrt{(z+x)(z+y)}\ge \frac{9}{2}(xy+yz+zx)-\frac{9xyz}{2(x+y+z)}\ge 4(xy+yz+zx).$$ The equality occurs when $x=y=z$.

---

Alternative solution.

Employing the Cauchy-Schwartz inequality we get $$(x+y)\sqrt{(z+x)(z+y)}=\sqrt{(x^2+xy+yz+zx)(y^2+xy+yz+zx)}\ge xy+(xy+yz+zx).$$ Therefore $$\sum _{\text{cyc}}(x+y)\sqrt{(z+x)(z+y)}\ge 4(xy+yz+zx).$$