Inequality In Triangle: Sides and Angle Bisectors

Problem

inequality in triangle X

Proof 1

We'll appeal to the AM-GM inequality, the Cauchy-Schwarz inequality, an expression for angles bisectors and the well known $\cos^2A+\cos^2B+\cos^2C\le\displaystyle\frac{9}{4}.$

$\displaystyle\begin{align} \sum_{cycl}l_a &= \sum_{cycl}\frac{bc\cos\displaystyle\frac{A}{2}}{b+c}\\ &\le 2\sqrt{\sum_{cycl}\left(\frac{bc}{b+c}\right)^2}\cdot\sqrt{\sum_{cycl}\cos^2\frac{A}{2}}\\ &\le 2\frac{3}{2}\sqrt{\sum_{cycl}\left(\frac{bc}{2\sqrt{bc}}\right)^2}\\ &\le\frac{3}{2}\sqrt{\sum_{cycl}\sqrt{bc}}\\ &\le\frac{3}{2}\frac{\displaystyle\sum_{cycl}a}{\sqrt{3}}. \end{align}$

Proof 2

This proof is used about the same basic tools but starts with a different formula for angle bisectors.

$\displaystyle\begin{align} \sqrt{3}l_a &= \frac{2\sqrt{bc}\sqrt{3p(p-a)}}{b+c}\\ &\le\frac{\displaystyle (b+c)\frac{3(p-a)+p}{2}}{b+c}\\ &=2p-\frac{3}{2}a\\ &=(b+c)-\frac{1}{2}a, \end{align}$

where $p=\frac{1}{2}(a+b+c)\;$ is the semiperimeter of $\Delta ABC.\;$ We conclude that $\displaystyle\frac{a}{2}+\sqrt{3}l_a\le b+c.\;$ There are two similar inequalities whose sum proves the result.

Acknowledgment

The inequality and two proofs have been posted by Dao Thanh Oai at the CutTheKnotMath facebook page. Proof 1 is by Ngo Dinh Tuan (Vietnam); Proof 2 is by Tran Trung Hieu (Vietnam).

 

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