MathDB
a+b+c=xyz=1 inequality in 6 variables

Source: Moldova JTST 2024 P1

June 10, 2024
inequalities

Problem Statement

Let a,b,c,x,y,za,b,c,x,y,z be positive real numbers, such that a+b+c=xyz=1a+b+c=xyz=1 Prove that: x23a+2+y23b+2+z23c+21 \frac{x^2}{3a+2}+\frac{y^2}{3b+2}+\frac{z^2}{3c+2} \ge 1 When does equality hold?