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 2 of 5: Write the three decompositions
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x2+yz2x2(y+z)=y+z2+(x−y)(x−z)2x2(y+z)\frac{x^2+yz}{\sqrt{2x^2(y+z)}}=\sqrt{\frac{y+z}{2}}+\frac{(x-y)(x-z)}{\sqrt{2x^2(y+z)}}
Detailed analysis

For example, x(y+z)2x2(y+z)=y+z2\frac{x(y+z)}{\sqrt{2x^2(y+z)}}=\sqrt{\frac{y+z}{2}}. Thus the full left side equals ∑cycy+z2\sum_{cyc}\sqrt{\frac{y+z}{2}} plus the three residual terms displayed by the cyclic version of this identity.