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 1 of 5: Split each numerator
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x2+yz=x(y+z)+(x−y)(x−z)x^2+yz=x(y+z)+(x-y)(x-z)
Detailed analysis

Use x2+yz=x(y+z)+(x−y)(x−z)x^2+yz=x(y+z)+(x-y)(x-z) and its cyclic analogues. Since x,y,z>0x,y,z>0, each cyclic term becomes a square-root part plus a residual difference quotient.