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I was trying to prove Nesbitt inequality using symmetry. So I assumed that $$a+b=x , b+c =y ,c+a=z$$

and then I rewrite inequality as:$$\sum \limits_{cyc} \frac {\frac {x+z-y}{2}}{y}\geq \frac{3}{2}$$

multiply both sides by $2$:$$\sum \limits_{cyc} \frac {x+z-y}{y}\geq 3$$

and adding $3$ to both sides:$$\sum \limits_{cyc} \frac {x+z}{y}\geq 6$$

And I stuck here.I searched on google and only source I found that solved this inequality like that was Wikipedia. And Wiki mentioned that $\sum \limits_{cyc} \frac {x+z}{y}\geq 6$ is true by am-gm, but I can't figure out how.

Martin Sleziak
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user2838619
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1 Answers1

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Hint: There are six terms on expansion of $\displaystyle \sum_{cyc} \frac{x+z}y = \sum_{sym} \frac{x}y$, just do AM-GM on the whole lot!

Macavity
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