olymmath

CIIM 2012 Problem 3

  • Day 1

Let a,b,c,a,b,c, the lengths of the sides of a triangle. Prove that (3a+b)(3b+a)(2a+c)(2b+c)+(3b+c)(3c+b)(2b+a)(2c+a)+(3c+a)(3a+c)(2c+b)(2a+b)≥4.\sqrt{\frac{(3a+b)(3b+a)}{(2a+c)(2b+c)}} + \sqrt{\frac{(3b+c)(3c+b)}{(2b+a)(2c+a)}} + \sqrt{\frac{(3c+a)(3a+c)}{(2c+b)(2a+b)}} \geq 4.

Dataset text AI-MO AoPS. Quoted with source.

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