# An Inequality with Sides, Altitudes, Angle Bisectors and Medians

### Solution

Note that the condition $c\le b\le a\,$ implies $(a-c)(b-c)(a-b)\ge 0.\,$ Using that we prove that

$\displaystyle \frac{b}{a}+\frac{c}{b}+\frac{a}{c}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}.$

The two are equivalent as follows from the sequence below:

\displaystyle \begin{align} &\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}&\Longleftrightarrow\\ &b^2c+c^2a+a^2b\ge a^2c+b^2a+c^2b&\Longleftrightarrow\\ &b^2(c-a)+ca(c-a)-b(c^2-a^2)\ge 0&\Longleftrightarrow\\ &(c-a)(b^2+ca-bc-ba)\ge 0&\Longleftrightarrow\\ &(c-a)(b(b-c)+a(c-b))\ge 0&\Longleftrightarrow\\ &(c-a)(b-c)(a-b)\ge 0. \end{align}

To continue, since, say, $h_a\le\ell_a\le m_a,$

\displaystyle\begin{align} \frac{b}{a}+\frac{c}{b}+\frac{a}{c} &\ge \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\\ &\ge \frac{h_b}{h_a}+\frac{h_c}{h_b}+\frac{h_a}{h_c}\\ &\ge \frac{h_b}{\ell_a}+\frac{h_c}{\ell_b}+\frac{h_a}{\ell_c}\\ &\ge \frac{h_b}{m_a}+\frac{h_c}{m_b}+\frac{h_a}{m_c}. \end{align}

Thus we have

$\displaystyle\frac{h_b}{m_a}+\frac{h_c}{m_b}+\frac{h_a}{m_c}\le \frac{b}{a}+\frac{c}{b}+\frac{a}{c}$

and also

$\displaystyle\frac{h_b}{\ell_a}+\frac{h_c}{\ell_b}+\frac{h_a}{\ell_c}\le \frac{b}{a}+\frac{c}{b}+\frac{a}{c}.$

Taking the product of the two gives

$\displaystyle \left(\frac{h_b}{m_a}+\frac{h_c}{m_b}+\frac{h_a}{m_c}\right)\left(\frac{h_b}{\ell_a}+\frac{h_c}{\ell_b}+\frac{h_a}{\ell_c}\right)\le \left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)^2,$

as required.

### Acknowledgment

The problem (from the Romanian Mathematical Magazine) has been kindly posted by Dan Sitaru at the CutTheKnotMath facebook page, Dan later emailed me his solution in a LaTeX file.