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sum (x - yz)^2/(x + yz)^2 <1 if x,y,z>0 with x+y+z = 1, x> yz, y > zx, z > xy

Source: (2021-) 2022 XV 15th Dürer Math Competition Finals Day 1 E+3

November 29, 2022
algebrainequalities

Problem Statement

Let x,y,zx, y, z denote positive real numbers for which x+y+z=1x+y+z = 1 and x>yzx > yz, y>zxy > zx, z>xyz > xy. Prove that (x−yzx+yz)2+(y−zxy+zx)2+(z−xyz+xy)2<1.\left(\frac{x - yz}{x + yz}\right)^2+ \left(\frac{y - zx}{y + zx}\right)^2+\left(\frac{z - xy}{z + xy}\right)^2< 1.
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