Let $\ \displaystyle x,y,z,x',y',z'$ positive real numbers such that $$xy+xz+yz=1$$ $$x'y'+x'z'+y'z'=1$$ Prove that:
$$xx'+yy'+zz'\geq 1$$
2026-04-04 04:34:42.1775277282
Inequality on six variables
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It's wrong!
Try $x=y'\rightarrow0^+$, $z=z'=\frac{1}{2}$ and $y=x'\rightarrow2$.