Problem
ALG-B2-M05-P020 Power product
#20
★★★★★ Level 5 of 5
Let \(x,y,z>0\) and \(x+y+z=3\). Prove \[x^x y^y z^z\ge1.\]
Hint. Take logarithms and consider \(f(t)=t\ln t\).
The function \(f(t)=t\ln t\) is convex for \(t>0\). Jensen gives \[\frac{x\ln x+y\ln y+z\ln z}{3}\ge f\left(\frac{x+y+z}{3}\right)=f(1)=0.\] Therefore \(\ln(x^x y^y z^z)\ge0\), so \(x^x y^y z^z\ge1\).
An olympiad-style function-choice problem: \(x^x\) should suggest taking logarithms.