Maths Olympiad Prep

Track / Stage 6 / 286 of 400 #1286 of 1964

Problem 1286

National olympiad, first round
Algebra Difficulty 6.5 Prove it

50. Let x,y,zx, y, z be positive numbers, and x+y=z=1x+y=z=1, prove that: xyz+xy+yzx+yz+zxy+zx⩽32⋅\sqrt{\frac{x y}{z+x y}}+\sqrt{\frac{y z}{x+y z}}+\sqrt{\frac{z x}{y+z x}} \leqslant \frac{3}{2} \cdot(2010 Kyrgyzstan Mathematical Olympiad Problem)

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

50. Since x+y+z=1x+y+z=1, we have
xyz+xy=xyz(x+y+z)+xy=xy(x+z)(y+z)\sqrt{\frac{x y}{z+x y}}=\sqrt{\frac{x y}{z(x+y+z)+x y}}=\sqrt{\frac{x y}{(x+z)(y+z)}}

Therefore, by the Cauchy-Schwarz inequality and using (x+y)(y+z)(z+x)⩾8xyz(x+y)(y+z)(z+x) \geqslant 8 x y z, we get
xyz+xy+yzx+yz+zxy+zx=xy(x+y)+yz(y+z)+zx(z+x)(x+y)(y+z)(z+x)⩽(xy+yz+zx)[(x+y)+(y+z)+(z+x)](x+y)(y+z)(z+x)=2(xy+yz+zx)(x+y+z)(x+y)(y+z)(z+x)=21+xyz(x+y)(y+z)(z+x)⩽21+18=32\begin{array}{l} \sqrt{\frac{x y}{z+x y}}+\sqrt{\frac{y z}{x+y z}}+\sqrt{\frac{z x}{y+z x}}=\frac{\sqrt{x y(x+y)}+\sqrt{y z(y+z)}+\sqrt{z x(z+x)}}{\sqrt{(x+y)(y+z)(z+x)}} \leqslant \\ \frac{\sqrt{(x y+y z+z x)[(x+y)+(y+z)+(z+x)]}}{\sqrt{(x+y)(y+z)(z+x)}}= \\ \sqrt{2} \frac{\sqrt{(x y+y z+z x)(x+y+z)}}{\sqrt{(x+y)(y+z)(z+x)}}=\sqrt{2} \sqrt{1+\frac{x y z}{(x+y)(y+z)(z+x)}} \leqslant \\ \sqrt{2} \sqrt{1+\frac{1}{8}}=\frac{3}{2} \end{array}

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.