An Inequality in Acute Triangle, Courtesy of Ceva's Theorem


inequality in triangle, Courtesy of Ceva's Theorem


Assume that in an acute $\Delta ABC,\;$ $AA_0,\;$ $BB_0,\;$ $CC_0\;$ are concurrent cevians. Then

$8\cdot BA_0\cdot CB_0\cdot AC_0\le abc.$

For convenience, denote $BA_0=x_1,\;$ $A_0C=y_1,\;$ $CB_0=x_2,\;$ $B_0A=y_2,\;$ $AC_0=x_3,\;$ $C_0B=y_3.\;$ Then, by Ceva's theorem,

$\displaystyle\frac{x_1}{y_1}\cdot \frac{x_2}{y_2}\cdot \frac{x_3}{y_3} = 1.$

We have to prove that $8x_1x_2x_3\le (x_1+y_1)(x_2+y_2)(x_3+y_3).\;$ In other words, we need to show that

$\displaystyle \frac{x_1+y_1}{x_1}\cdot\frac{x_2+y_2}{x_2}\cdot\frac{x_3+y_3}{x_3}\ge 8,$


$\displaystyle \left(1+\frac{y_1}{x_1}\right)\cdot\left(1+\frac{y_2}{x_2}\right)\cdot\left(1+\frac{y_3}{x_3}\right)\ge 8,$

Multiplying out, this is reduced to

$\displaystyle 1+\frac{y_1}{x_1}+\frac{y_2}{x_2}+\frac{y_3}{x_3}+ +\frac{y_1y_2}{x_1x_2}+\frac{y_2y_3}{x_2x_3}+\frac{y_3y_1}{x_3x_1}+\frac{y_1y_2y_3}{x_1x_2x_3}\ge 8.$

Making multiple uses of Ceva's theorem, this is equivalent to

$\displaystyle 1+\frac{y_1}{x_1}+\frac{y_2}{x_2}+\frac{y_3}{x_3}+ +\frac{x_3}{y_3}+\frac{x_1}{y_1}+\frac{x_2}{y_2}+1\ge 8$

and, in turn, to

$\displaystyle \left(\frac{y_1}{x_1}+\frac{x_1}{y_1}\right)+\left(\frac{y_2}{x_2}++\frac{x_2}{y_2}\right)+\left(\frac{y_3}{x_3}+ +\frac{x_3}{y_3}\right)\ge 6$

which is true by the AM-GM inequality applied thrice.


Since the triples of angle bisectors, altitudes, and symmedians are all concurrent cevians, we may apply the lemma to each triple:

$8\cdot AB'\cdot BC'\cdot CA'\le abc\\ 8\cdot AB''\cdot BC''\cdot CA''\le abc\\ 8\cdot AB'''\cdot BC'''\cdot CA'''\le abc.$

Adding up gives the desired result.


The inequality with the solution has been posted by Dan Sitaru at the CutTheKnotMath facebook page.


|Contact| |Front page| |Contents| |Inequalities| |Geometry| |Up|

Copyright © 1996-2018 Alexander Bogomolny


Search by google: