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Berkeley Math Circle: Monthly Contest 6

Algebra Difficulty 4.4 AIME Prove it United States

Problem:
Prove that if x,y,zx, y, z are positive real numbers, then
x2+2y2+3z2>xy+3yz+zx. x^{2}+2 y^{2}+3 z^{2}>x y+3 y z+z x .

Solution

Solution:
We note the identity
2(x2+2y2+3z2)−2(xy+3yz+zx)=(x−y)2+(x−z)2+3(y−z)2+2z2≥2z2>0 2\left(x^{2}+2 y^{2}+3 z^{2}\right)-2(x y+3 y z+z x)=(x-y)^{2}+(x-z)^{2}+3(y-z)^{2}+2 z^{2} \geq 2 z^{2}>0
since all squares of real numbers are nonnegative.

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