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1/(x + y + z)<= (x + y)(\sqrt3 z + 1) if x,y,z>=0 with xy + yz + zx = 1

Source: Switzerland - 2017 Swiss MO Final Round p10

January 14, 2023
algebrainequalities

Problem Statement

Let x,y,zx, y, z be nonnegative real numbers with xy+yz+zx=1xy + yz + zx = 1. Show that: 4x+y+z≤(x+y)(3z+1).\frac{4}{x + y + z} \le (x + y)(\sqrt3 z + 1).
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