Maths Olympiad Prep

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Algebra Difficulty 5.6 AIME, harder Find the answer

109. Find all real numbers x,y,z,wx, y, z, w such that
x+y+z=32,4x1+4y1+4z12+3w2x+y+z=\frac{3}{2}, \sqrt{4 x-1}+\sqrt{4 y-1}+\sqrt{4 z-1} \geqslant 2+3 \sqrt{w-2}
(2007-2008 Hungarian Mathematical Olympiad)

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

109. By the mean inequality,
4x1+4y1+4z134x1+4y1+4z13=1\frac{\sqrt{4 x-1}+\sqrt{4 y-1}+\sqrt{4 z-1}}{3} \leqslant \sqrt{\frac{4 x-1+4 y-1+4 z-1}{3}}=1

Equality holds if and only if x=y=z=12x=y=z=\frac{1}{2}. Therefore, 2+3w24x1+2+3 \sqrt{w-2} \leqslant \sqrt{4 x-1}+ 4y1+4z13\sqrt{4 y-1}+\sqrt{4 z-1} \leqslant 3, hence, w20\sqrt{w-2} \leqslant 0. Thus, w=2w=2, and all the equalities hold. At this point, x=y=z=12,w=2x=y=z=\frac{1}{2}, w=2.

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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.