Fermat_Theorem wrote: If $a,b,c$ are positive numbers such that $abc = 1$, prove the inequality $\frac{1}{\sqrt{b+\frac{1}{a}+\frac{1}{2}}} + \frac{1}{\sqrt{c+\frac{1}{b}+\frac{1}{2}}} + \frac{1}{\sqrt{a+\frac{1}{c}+\frac{1}{2}}} \geq \sqrt{2}$ The problem is obviously something else because the equality case $a=b=c=1$ doesn't even work

Fermat_Theorem wrote: If $a,b,c$ are positive numbers such that $abc = 1$, prove the inequality $\frac{1}{\sqrt{b+\frac{1}{a}+\frac{1}{2}}} + \frac{1}{\sqrt{c+\frac{1}{b}+\frac{1}{2}}} + \frac{1}{\sqrt{a+\frac{1}{c}+\frac{1}{2}}} \geq \sqrt{2}$ Moroccan tst 2006

From $AM-GM$ inequality $1+b+\frac{1}{a}=\frac{1}{2}+b+\frac{1}{a}+\frac{1}{2} \geq 2\sqrt{\frac{1}{2}(b+\frac{1}{a}+\frac{1}{2})} \implies \frac{1}{\sqrt{b+\frac{1}{a}+\frac{1}{2}}} \geq \frac{\sqrt{2}}{1+b+\frac{1}{a}}$ Similarly: $\frac{1}{\sqrt{c+\frac{1}{b}+\frac{1}{2}}} \geq \frac{\sqrt{2}}{1+c+\frac{1}{b}}$ and $\frac{1}{\sqrt{a+\frac{1}{c}+\frac{1}{2}}} \geq \frac{\sqrt{2}}{1+a+\frac{1}{c}}$. We will now add up these inequalities to get: $\frac{1}{\sqrt{b+\frac{1}{a}+\frac{1}{2}}} + \frac{1}{\sqrt{c+\frac{1}{b}+\frac{1}{2}}} + \frac{1}{\sqrt{a+\frac{1}{c}+\frac{1}{2}}} \geq \sqrt{2}(\frac{1}{1+b+\frac{1}{a}}+\frac{1}{1+c+\frac{1}{b}}+\frac{1}{1+a+\frac{1}{c}})$ After long factorization and using that $abc=1$ we get that $\frac{1}{1+b+\frac{1}{a}}+\frac{1}{1+c+\frac{1}{b}}+\frac{1}{1+a+\frac{1}{c}}=\frac{6+3ab+3a+3ac+a^2c+3c+3bc+3b+ab^2+bc^2}{6+3ab+3a+3ac+a^2c+3c+3bc+3b+ab^2+bc^2}=1$ And now finally: $\frac{1}{\sqrt{b+\frac{1}{a}+\frac{1}{2}}} + \frac{1}{\sqrt{c+\frac{1}{b}+\frac{1}{2}}} + \frac{1}{\sqrt{a+\frac{1}{c}+\frac{1}{2}}} \geq \sqrt{2}$