For non-negative $x,y,z$ satisfy $\frac{1}{2x^2+1}+\frac{1}{2y^2+1}+\frac{1}{2z^2+1}=1$ then show that $x^2+y^2+z^2+6\geq 3(x+y+z)$
Idea how to handle the constraint? I'm unaware .
2026-03-25 09:24:30.1774430670
Inequality on non-negative reals
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Yes, it's true for all reals.
$\sum\limits_{cyc}(x^2-3x+2)=\sum\limits_{cyc}\left(x^2-3x+2-\frac{9}{4}\left(\frac{1}{2x^2+1}-\frac{1}{3}\right)\right)=\sum\limits_{cyc}\frac{(x-1)^2(2x-1)^2}{2(2x^2+1)}\geq0$.
Done!