Problem 4

Let

x,y,z

be positive real numbers such that

x+y+z=1

. Prove that

\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 3 of 5: Show the residual sum is nonnegative

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x\ge y\ge z\Longrightarrow R_x\ge0,\quad R_z+R_y\ge (y-z)(x-y)\left(\frac1{\sqrt{2z^2(x+y)}}-\frac1{\sqrt{2y^2(z+x)}}\right)\ge0

Detailed analysis

By symmetry assume $x\ge y\ge z$ . The $x$ residual is nonnegative. Pair the other two residuals: replacing $x-z$ by the smaller $x-y$ gives $R_y+R_z\ge (y-z)(x-y)\left(\frac1{\sqrt{2z^2(x+y)}}-\frac1{\sqrt{2y^2(z+x)}}\right)$ . The last bracket is nonnegative because $y\ge z$ and $y^2(x+z)\ge z^2(x+y)$ .