### 2006 IMO SL #A5

If $a,b,c$ are the sides of a triangle, prove that $\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3$

Let $x=\sqrt{b}+\sqrt{c}-\sqrt{a}$ and similarly define $y, z$. Note $x,y,z>0$. Then, $$0 < b+c-a = \left( \frac{x+y}{2} \right)^2 + \left( \frac{x+z}{2} \right)^2 - \left( \frac{y+z}{2} \right)^2$$ $$= x^2+xy+xz-yz.$$ So, we have to show $\sum_{\text{cyc}} \frac{\sqrt{x^2+xy+xz-yz}}{x} \le 3\sqrt 2.$

Cauchy gives $\left( \sum_{\text{cyc}} \frac{\sqrt{x^2+xy+xz-yz}}{x} \right)^2 \le 3 \sum_{\text{cyc}} \frac{x^2+xy+xz-yz}{x^2},$ from which we can conclude that it suffices to prove $\sum_{\text{sym}} xy^2z^3 \le (xy)^3 + (yz)^3 + (zx)^3 + 3(xyz)^2.$ But this easily follows from Schur's inequality on $xy, yz, zx$, so we are done. $\square$