Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 7.5 National olympiad, round 2 Prove it

Example 7 (Neuberg-Pedoe Inequality) Let the side lengths of $\triangle A_{1} A_{2} A_{3}$ and $\triangle B_{1} B_{2} B_{3}$ be $a_{1}, a_{2}, a_{3}$ and $b_{1}, b_{2}, b_{3}$ , respectively, and their areas be denoted as $S_{1}$ and $S_{2}$ . Prove:
$\begin{aligned} & a_{1}^{2}\left(b_{2}^{2}+b_{3}^{2}-b_{1}^{2}\right)+a_{2}^{2}\left(b_{3}^{2}+b_{1}^{2}-b_{2}^{2}\right)+a_{3}^{2}\left(b_{1}^{2}+b_{2}^{2}-b_{3}^{2}\right) \\ \geqslant & 16 S_{1} S_{2}, \end{aligned}$

with equality if and only if $\triangle A_{1} A_{2} A_{3} \backsim \triangle B_{1} B_{2} B_{3}$ .

Solution

Prove that by slightly transforming equation (13), we can obtain its equivalent form:
$16 S_{1} S_{2} \leqslant\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)-2\left(a_{1}^{2} b_{1}^{2}+a_{2}^{2} b_{2}^{2}+a_{3}^{2} b_{3}^{2}\right) .$

Rearranging terms and applying the Cauchy-Schwarz inequality, we get
$\begin{aligned} & 16 S_{1} S_{2}+2\left(a_{1}^{2} b_{1}^{2}+a_{2}^{2} b_{2}^{2}+a_{3}^{2} b_{3}^{2}\right) \\ \leqslant & \sqrt{\left(16 S_{1}^{2}+2\left(a_{1}^{4}+a_{2}^{4}+a_{3}^{4}\right)\right)\left(16 S_{2}^{2}+2\left(b_{1}^{4}+b_{2}^{4}+b_{3}^{4}\right)\right)} \\ = & \left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right) \end{aligned}$

Equality holds if and only if $S_{1}: S_{2}=a_{1}^{2}: b_{1}^{2}=a_{2}^{2}: b_{2}^{2}=a_{3}^{2}: b_{3}^{2}$ , i.e., when $\triangle A_{1} A_{2} A_{3} \leadsto \triangle B_{1} B_{2} B_{3}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.