old and easy: http://www.mathlinks.ro/Forum/viewtopic.php?t=14765 http://www.mathlinks.ro/Forum/viewtopic.php?t=32861 dg

k2c901_1 wrote: Positive reals $a,b,c$ satisfy $abc=1$. Prove that $ 1+\frac{3}{a+b+c}\ge \frac{6}{ab+bc+ca}$. We have: \[\left(1-\sqrt{3 \over a+b+c}\right)^{2}\ge 0 \\ \implies 1+{3 \over a+b+c}\ge{6 \over \sqrt{3(a+b+c)}}={6 \over \sqrt{3abc(a+b+c)}}\ge{6 \over ab+bc+ca}.\]

Let $a,$ $b$ and $c$ are positive numbers such that $abc=1$. Prove that $1+\frac{8}{a+b+c}\ge \frac{11}{ab+bc+ca}$ Is it true?

The stronger inequality holds: $ 1+\frac{7}{a+b+c}\ge \frac{10}{ab+bc+ca}$.

wrong

silouan wrote: But $xy+yz+zx\leq \sqrt{3(x+y+z)}$ But $xy+yz+zx\geq \sqrt{3(x+y+z)}$

Sorry I was wrong