Balkan Mathematical Olympiad Shortlisted Problems

Lemma. For positive $a$, $b$, $c$, $x$, $y$, $z$, it holds that $ax+by+cz+2\sqrt{ab+bc+ca}\sqrt{xy+yz+zx}\le (a+b+c)(x+y+z)$

Proof. Use Cauchy-Schwarz inequality $$\begin{array}{rl}ax+by+cz+2\sqrt{ab+bc+ca}\sqrt{xy+yz+zx}& \le \sqrt{a^2+b^2+c^2}\sqrt{x^2+y^2+z^2}+\sqrt{ab+bc+ca}\sqrt{xy+yz+zx}+\sqrt{ab+bc+ca}\sqrt{xy+yz+zx}\\ & \le \sqrt{a^2+b^2+c^2+2ab+2bc+2ca}\sqrt{x^2+y^2+z^2+2xy+2yz+2zx}=(a+b+c)(x+y+z)\end{array}$$ Take $(x,y,z)=(\frac{a}{b+c},\frac{b}{a+c},\frac{c}{a+b})$ in the lemma. First we observe $$\sqrt{xy+yz+zx}=\sqrt{\frac{a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b}{(a+b)(b+c)(c+a)}}=\frac{1}{\sqrt{1+2abc}}$$ By the lemma, we have $$2\sqrt{ab+bc+ca}\sqrt{xy+yz+zx}\le (a+b+c)(x+y+z)-ax-by-cz$$ $$\frac{2\sqrt{ab+bc+ca}}{\sqrt{1+2abc}}\le ay+az+bx+bz+cx+cy$$ $$\frac{2\sqrt{ab+bc+ca}}{\sqrt{1+2abc}}\le x(b+c)+y(c+a)+z(a+b)=a+b+c$$ And finally, squaring both sides, we get the desired inequality $$\frac{ab+bc+ca}{1+2abc}\le \frac{(a+b+c{)}^2}{4}$$