Hellenic Mathematical Olympiad

(A) It is enough to prove $$\begin{array}{rl}& {\kappa }^2+{\lambda }^2+{\mu }^2+2\kappa \lambda +2\lambda \mu +2\mu \kappa \ge 3\kappa \lambda +3\lambda \mu +3\mu \kappa \\ \iff & {\kappa }^2+{\lambda }^2+{\mu }^2-\kappa \lambda -\lambda \mu -\mu \kappa \ge 0\\ \iff & (\kappa -\lambda {)}^2+(\lambda -\mu {)}^2+(\mu -\kappa {)}^2\ge 0,\end{array}$$ which is valid for all real numbers $\kappa ,\lambda ,\mu$. Equality holds if and only if $\kappa =\lambda =\mu$.

(B) From the given relation we get: $$\frac{1}{\alpha }=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\text{and}\frac{1}{\beta }=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}.$$ Thus we have $\alpha ,\beta >0$ and from question (A) by putting $\kappa =\frac{1}{x},\lambda =\frac{1}{y},\mu =\frac{1}{z}$, we get $$\frac{1}{{\beta }^2}={\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}^2\ge 3\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\Rightarrow \alpha \ge 3{\beta }^2.$$ Equality holds if and only if $x=y=z$.