# An Inequality in Triangle with Altitudes, Medians And Symmedians

### Solution

Let $BA_3=m\,$ and $CA_3=n.\,$ Then $\displaystyle\frac{n}{m}=\frac{b^2}{c^2}\,$ implying $\displaystyle\frac{n+m}{m}=\frac{b^2+c^2}{c^2},\,$ i.e., $\displaystyle\frac{a}{m}=\frac{b^2+c^2}{c^2},\,$ and, subsequently, $\displaystyle m=\frac{ac^2}{b^2+c^2},\,$ so $\displaystyle BA_3=\frac{ac^2}{b^2+c^2}.\,$ It follows that

\displaystyle\begin{align} A_2A_3&=BA_2-BA_3=\frac{a}{2}-\frac{ac^2}{b^2+c^2}\\ &=\frac{a(b^2-c^2)}{2(b^2+c^2)}. \end{align}

\displaystyle\begin{align} A_1A_2&=BA_2-BA_1=\frac{a}{2}-c\cos B=\frac{a}{2}-\frac{c(c^2+a^2-b^2)}{2ca}\\ &=\frac{a}{2}-\frac{c^2+a^2-b^2}{2a}\\ &=\frac{b^2-c^2}{2a}. \end{align}

Thus,

$\displaystyle\frac{A_2A_3}{A_1A_2}=\frac{a(b^2-c^2)}{2(a^2+b^2)}\cdot\frac{2a}{b^2-c^2}=\frac{a^2}{b^2+c^2}.$

Similarly, $\displaystyle\frac{B_2B_3}{B_1B_2}=\frac{b^2}{c^2+a^2}\,$ and $\displaystyle\frac{C_2C_3}{C_1C_2}=\frac{c^2}{a^2+b^2}.\,$ So, by using Padoa's inequality, we have

\displaystyle\begin{align} LHS &= \frac{a^2b^2c^2}{(b^2+c^2)(c^2+a^2)(a^2+b^2)}\\ &\gt\frac{\displaystyle\left(\prod_{cycl}(a+b-c)\right)^2}{(b^2+c^2)(c^2+a^2)(a^2+b^2)}\\ &\gt\frac{\displaystyle\left(\prod_{cycl}(a+b-c)\right)^2}{(2b^2+2c^2-a^2)(2c^2+2a^2-b^2)(2a^2+2b^2-c^2)}\\ &=\prod_{cycl}\frac{(a+b-c)^2}{2a^2+2b^2-c^2}. \end{align}

(Note that $b^2+c^2\gt a^2\,$ so that $2b^2+2c^2-a^2\gt b^2+c^2.)$

### Acknowledgment

Dan Sitaru has kindly posted at the CutTheKnotMath facebook page the above problem of his from the Romanian Mathematical Magazine, with a solution by Soumava Chakraborty.