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Berkeley Math Circle Monthly Contest 3

Algebra Difficulty 4.8 AIME Prove it United States

Problem:

Let xx, yy, and zz be real numbers such that xyz=1x y z = 1. Prove that
x2+y2+z2≥1x+1y+1z x^{2} + y^{2} + z^{2} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z}

Solution

Solution:

Replacing the 11 in the numerators of the fractions on the right by xyzx y z, it suffices to prove that
x2+y2+z2≥yz+zx+xy x^{2} + y^{2} + z^{2} \geq y z + z x + x y
which is true because
x2+y2+z2−yz−zx−xy=(x−y)2+(y−z)2+(z−x)22≥0. x^{2} + y^{2} + z^{2} - y z - z x - x y = \frac{(x - y)^{2} + (y - z)^{2} + (z - x)^{2}}{2} \geq 0.

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