Maths Olympiad Prep

Track / Stage 5 / 93 of 400 #693 of 1964

Problem 693

AIME late
Algebra Difficulty 5.1 Prove it Korean Mathematical Olympiad Final Round · South Korea

Let aa, bb and cc be the sides of a triangle, and we set
A=a2+bcb+c+b2+cac+a+c2+aba+b A = \frac{a^2 + bc}{b+c} + \frac{b^2 + ca}{c+a} + \frac{c^2 + ab}{a+b}
B=1(a+b−c)(b+c−a)+1(b+c−a)(c+a−b)+1(c+a−b)(a+b−c) B = \frac{1}{\sqrt{(a+b-c)(b+c-a)}} + \frac{1}{\sqrt{(b+c-a)(c+a-b)}} + \frac{1}{\sqrt{(c+a-b)(a+b-c)}}
Prove that AB≥9AB \ge 9.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

Clearly
B≥1b+1c+1a B \ge \frac{1}{b} + \frac{1}{c} + \frac{1}{a}
and
A−(a+b+c)=a4+b4+c4−a2b2−b2c2−c2a2(a+b)(b+c)(c+a)≥0. A - (a + b + c) = \frac{a^4 + b^4 + c^4 - a^2b^2 - b^2c^2 - c^2a^2}{(a+b)(b+c)(c+a)} \ge 0.
Therefore we have
AB≥(1b+1c+1a)(b+c+a)≥9. AB \ge \left( \frac{1}{b} + \frac{1}{c} + \frac{1}{a} \right) (b+c+a) \ge 9.
by Cauchy-Schwarz inequality. This completes the proof. □

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.