Let x,y,z be positive real numbers such that x+y+z=1. Find the minimum value of x+y1+x+z1+y+z1.
Solution — click to reveal
By Cauchy-Schwarz, [(x+y)+(x+z)+(y+z)](x+y1+x+z1+y+z1)≥(1+1+1)2=9,so x+y1+x+z1+y+z1≥2(x+y+z)9=29.Equality occurs when x=y=z=31, so the minimum value is 29.