SELECTION EXAMINATION 2019

By putting $x=\frac{1}{\alpha }$, $y=\frac{1}{\beta }$, $z=\frac{1}{\gamma }$, we have $x+y+z=3$ and the inequality takes the form: $$\frac{xy(x+y)}{x^2+xy+y^2}+\frac{yz(y+z)}{y^2+yz+z^2}+\frac{zx(z+x)}{z^2+zx+x^2}\le 2.$$ We have $\frac{xy}{x^2+xy+y^2}\le \frac{1}{3}\iff x^2+xy+y^2\ge 3xy\iff (x-y{)}^2\ge 0$ and equality is valid when $x=y$. Multiplying both sides by $x+y>0$, we get $$\frac{xy(x+y)}{x^2+xy+y^2}\le \frac{1}{3}(x+y).(1)$$ Similarly we get the relations: $$\frac{yz(y+z)}{y^2+yz+z^2}\le \frac{1}{3}(y+z)(2),\frac{zx(z+x)}{z^2+zx+x^2}\le \frac{1}{3}(z+x).(3)$$ Finally using summation of (1), (2) and (3) we have $$\frac{xy(x+y)}{x^2+xy+y^2}+\frac{yz(y+z)}{y^2+yz+z^2}+\frac{zx(z+x)}{z^2+zx+x^2}\le \frac{1}{3}\cdot 2(x+y+z)=2.$$ Equality is valid when $x=y=z=1$ or $\alpha =\beta =\gamma =1$.