Geometric Inequality problem

Question

Within a problem I was solving I came across

Prove $xy+yz+xz \geq 2xyz$ for all non-negative real numbers $x,y,z$.

I have attempted to prove this for some time now, but cannot find a solution that works. The closest I came was trying the AM GM inequality with xy, yz, and xz as my 3 terms, but obviously we need 2 terms.

Have you considered $x=y=z$ ? you get something slightly dubious. There's probably another condition — Asinomás, Jul 01 '21 at 03:13
well these are just the segments formed by the tangents of the inscribed circle in a triangle if that helps — Smartsav10, Jul 01 '21 at 03:15
Well, for $x=y=z=2$ we have $4+4+4=12<16=2\cdot2\cdot2\cdot2$. — g.kov, Jul 01 '21 at 04:34

score 1 · Answer 1 · answered Jul 01 '21 at 09:40

Prove \begin{align}xy+yz+xz \ge 2xyz \tag{1}\label{1}\end{align} for all non-negative real numbers $x,y,z$.

A primitive counterexample: for non-negative real numbers $x=y=z=2$ we have \begin{align} 4+4+4&=12<16=2\cdot2\cdot2\cdot2 , \end{align}
so the statement \eqref{1} is obviously false.

Geometric Inequality problem

1 Answers1