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Problem 4 vietnamese tst 2006

Source: Vietnam Team Selection Tests 2006, Day 2, Problem 1

April 18, 2006
inequalitiesfunctionconvex-concave inequalitiesthree variable inequality

Problem Statement

Prove that for all real numbers x,y,z∈[1,2]x,y,z \in [1,2] the following inequality always holds: (x+y+z)(1x+1y+1z)≥6(xy+z+yz+x+zx+y). (x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 6(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}). When does the equality occur?
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