MathDB
Simple inequality

Source: Austria Mathematical Olympiad 2021?

May 31, 2021
inequalitiesalgebra

Problem Statement

Let a,b,c0a,b,c\geq 0 and a+b+c=1.a+b+c=1. Prove thata2a+1+b3b+1+c6c+112.\frac{a}{2a+1}+\frac{b}{3b+1}+\frac{c}{6c+1}\leq \frac{1}{2}. (Marian Dinca)