Let \(a_{1}, \ldots, a_{5} > 0\) whose sum of squares is at least \(1.\) Prove that we have: $$\frac{a_{1}^{2}}{a_{2}+a_{3}+a_{4}} +
\frac{a_{2}^{2}}{a_{3}+a_{4}+a_{5}} +
\frac{a_{3}^{2}}{a_{4}+a_{5}+a_{1}}+
\frac{a_{4}^{2}}{a_{5}+a_{1}+a_{2}}+
\frac{a_{5}^{2}}{a_{1}+a_{2}+a_{3}} \geq \frac{\sqrt{5}}{3}.$$
Mathematics and Youth
