MathDB
(x+y+z)/(xy+yz+zx) <= 1+5/247} ( (x-y)^2+(y-z)^2+(z-x)^2 )

Source: 2006 VMEO III Shortlist SL A3 Vietnamese Mathematics e - Olympiad https://artofproblemsolving.com/community/c2461015_vmeo__viet

October 3, 2021
algebrainequalities

Problem Statement

For positive real numbers x,y,zx,y,z that satisfy xy+yz+zx+xyz=4 xy + yz + zx + xyz=4, prove that x+y+zxy+yz+zx1+5247((xy)2+(yz)2+(zx)2)\frac{x+y+z}{xy+yz+zx}\le 1+\frac{5}{247}\cdot \left( (x-y)^2+(y-z)^2+(z-x)^2\right)