MathLabs

Problem 4

Let x,y,zx,y,z be positive real numbers such that x+y+z=1x+y+z=1. Prove that x2+yz2x2(y+z)+y2+zx2y2(z+x)+z2+xy2z2(x+y)≥1\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\ge1.
Step 4 of 5: Bound the square-root part
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∑cycy+z2≥(y+z)+(z+x)+(x+y)2=1\sum_{cyc}\sqrt{\frac{y+z}{2}}\ge\sqrt{\frac{(y+z)+(z+x)+(x+y)}2}=1
Detailed analysis

Cauchy–Schwarz gives ∑cycy+z2≥(y+z)+(z+x)+(x+y)2\sum_{cyc}\sqrt{\frac{y+z}{2}}\ge\sqrt{\frac{(y+z)+(z+x)+(x+y)}2}. Since x+y+z=1x+y+z=1, the right side equals 11.