Problem
ALG-B2-M05-P018 Three denominators with two
#18
★★★★★ Level 5 of 5
Let \(x,y,z\ge0\) and \(x+y+z=3\). Prove \[\frac1{2+x}+\frac1{2+y}+\frac1{2+z}\ge1.\]
Hint. Consider \(f(t)=\frac1{2+t}\).
The function \(f(t)=\frac1{2+t}\) is convex for \(t\ge0\). Jensen gives \[\frac13\sum\frac1{2+x}\ge \frac1{2+(x+y+z)/3}=\frac13.\] Multiply by \(3\).
This problem trains recognition of a shifted reciprocal as a convex function.