# An Inequality Solved by Changing Appearances

### Solution

\displaystyle \begin{align} &\sum_{cycl}a^2\cdot\sum_{cycl}\frac{1}{x^2}+\frac{\displaystyle 2\sum_{cycl}ab\cdot\sum_{cycl}x}{xyz}\\ &\qquad=\frac{a^2}{x^2}+\frac{a^2}{y^2}+\frac{a^2}{z^2}+\frac{b^2}{x^2}+\frac{b^2}{y^2}+\frac{b^2}{z^2}+\frac{c^2}{x^2}+\frac{c^2}{y^2}+\frac{c^2}{z^2}\\ &\qquad+\frac{2ab}{xy}+\frac{2ab}{yz}+\frac{2ab}{zx}+\frac{2bc}{xy}+\frac{2bc}{yz}+\frac{2bc}{zx}+\frac{2ca}{xy}+\frac{2ca}{yz}+\frac{2ca}{zx}\\ &\qquad=\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+\frac{2ab}{xy}+\frac{2bc}{yz}+\frac{2ca}{zx}\\ &\qquad+\frac{a^2}{y^2}+\frac{b^2}{z^2}+\frac{c^2}{x^2}+\frac{2ab}{yz}+\frac{2bc}{zx}+\frac{2ca}{xy}\\ &\qquad+\frac{a^2}{z^2}+\frac{b^2}{x^2}+\frac{c^2}{y^2}+\frac{2ab}{zx}+\frac{2bc}{xy}+\frac{2ca}{yz}\\ &\qquad=\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)^2+\left(\frac{a}{y}+\frac{b}{z}+\frac{c}{x}\right)^2+\left(\frac{a}{z}+\frac{b}{x}+\frac{c}{y}\right)^2\\ &\qquad\ge 0. \end{align}

### Acknowledgment

Dan Sitaru has kindly reposted the above problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page. The above solution is by Ravi Prakash.