Reference: R. Todev, Nordic Mathematical Contests, 1987-2009.
Question:
Let a, b, and c be positive real numbers. Prove that 
Solution:
The arithmetic-geometric inequality yields
![3=3\sqrt[3]{\frac{a^{2}}{b^{2}}.\frac{b^{2}}{c^{2}}.\frac{c^{2}}{a^{2}}}\leq \frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}](https://s0.wp.com/latex.php?latex=3%3D3%5Csqrt%5B3%5D%7B%5Cfrac%7Ba%5E%7B2%7D%7D%7Bb%5E%7B2%7D%7D.%5Cfrac%7Bb%5E%7B2%7D%7D%7Bc%5E%7B2%7D%7D.%5Cfrac%7Bc%5E%7B2%7D%7D%7Ba%5E%7B2%7D%7D%7D%5Cleq+%5Cfrac%7Ba%5E%7B2%7D%7D%7Bb%5E%7B2%7D%7D%2B%5Cfrac%7Bb%5E%7B2%7D%7D%7Bc%5E%7B2%7D%7D%2B%5Cfrac%7Bc%5E%7B2%7D%7D%7Ba%5E%7B2%7D%7D&bg=ffffff&fg=111111&s=0 "3=3\sqrt[3]{\frac{a^{2}}{b^{2}}.\frac{b^{2}}{c^{2}}.\frac{c^{2}}{a^{2}}}\leq \frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}")
or 
On the other hand, the Cauchy-Schwarz inequality implies
We arrive at the inequality we desire by combining relations I and II. Hence, the proof. QED.
Cheers,
Nalin Pithwa.
