# An Inequality in Triangle, VII

### Proof 1

First,

\displaystyle\begin{align} \sum_{cycl}\frac{m_a^2}{m_b^2} &= \sum_{cycl}\frac{1}{2}\left(\frac{m_a^2}{m_b^2}+\frac{m_b^2}{m_c^2}\right)\\ &\ge\sum_{cycl}\sqrt{\frac{m_a^2}{m_b^2}\cdot\frac{m_b^2}{m_c^2}}\\ &=\sum_{cycl}\frac{m_a}{m_c}. \end{align}

It follows that

\displaystyle\begin{align} \left(\sum_{cycl}\frac{m_a^2}{m_b^2}\right)\left(\sum_{cycl}x^2\right)&+2\left(\sum_{cycl}\frac{m_a}{m_c}\right)\left(\sum_{cycl}xy\right)\\ &\ge\left(\sum_{cycl}\frac{m_a}{m_c}\right)\left(\sum_{cycl}x^2\right)+2\left(\sum_{cycl}\frac{m_a}{m_c}\right)\left(\sum_{cycl}xy\right)\\ &=\left(\sum_{cycl}\frac{m_a}{m_c}\right)\left(\sum_{cycl}x^2+2\sum_{cycl}xy\right)\\ &=\left(\sum_{cycl}\frac{m_a}{m_c}\right)\left(\sum_{cycl}x\right)^2\\ &\ge 0. \end{align}

### Proof 2

Let's set $\displaystyle m=\frac{m_a}{m_b},\;$ $\displaystyle n=\frac{m_b}{m_c},\;$ $\displaystyle p=\frac{m_c}{m_a}\;$ Obviously, $mnp=1.$

Further, since $\displaystyle m^2+n^2+p^2\ge mn+np+pm =\frac{1}{m}+\frac{1}{n}+\frac{1}{p},\;$

\displaystyle\begin{align} (m^2+n^2+p^2)(x^2+y^2+z^2)&+2\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)(mn+np+pm)\\ &=(x+y+z)^2(mn+np+pn)\\ &\ge 0. \end{align}

### Acknowledgment

The inequality has been kindly posted at the CutTheKnotMath facebook page by Dan Sitaru, along with a proof by Marian Dinca (Proof 1); Proof 2 is by .