blackdivcp

2022-10-12

Prove that $\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\le 1$
If $abc=1$ then
$\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\le 1$
I have tried AM-GM and C-S and can't seem to find a solution. What is the best way to prove it?

Do you have a similar question?

wespee0

Expert

Also we can use the following reasoning.
For positives $a$, $b$ and $c$ let $a={x}^{3}$, $b={y}^{3}$ and $c={z}^{3}$
Hence, $\sum _{cyc}\frac{1}{a+b+1}=\sum _{cyc}\frac{1}{{x}^{3}+{y}^{3}+xyz}\le \sum _{cyc}\frac{1}{{x}^{2}y+{y}^{2}x+xyz}=\sum _{cyc}\frac{z}{x+y+z}=1$,
but I think the first way is better.

Still Have Questions?

Krish Logan

Expert

It's wrong, of course.
For positives $a$, $b$ and $c$ it's $\sum _{cyc}\left({a}^{2}b+{a}^{2}c-2a\right)\ge 0$, which is Muirhead because $\left(2,1,0\right)\succ \left(\frac{5}{3},\frac{2}{3},\frac{2}{3}\right)$.

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