<s>[INST] How do I prove this inequality $3(a+b+c+1)\ge 4 \left( \sqrt{\frac{a^2+1}{a+1}}+\sqrt{\frac{b^2+1}{b+1}}+\sqrt{\frac{c^2+1}{c+1}} \right)$? If $a,b,c\gt 0$ and $abc=1$, how do I prove the following inequality $3(a+b+c+1)\ge 4 \left( \sqrt{\frac{a^2+1}{a+1}}+\sqrt{\frac{b^2+1}{b+1}}+\sqrt{\frac{c^2+1}{c+1}} \right)$?
My version:
\begin{gathered}
\sqrt{\frac{a^{2}+1}{a+1}} \leq \sqrt{1+\frac{a^{2}-a}{a+1}} \leq 1+\frac{1}{2} \frac{a^{2}-a}{a+1}=\frac{a}{2}+\frac{1}{a+1} \\
\Leftrightarrow 4\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right) \leq 3+(a+b+c), a b c=1 \\
a+b+c=3 t^{2}=p \geq 3, a b+b c+c a=q \geq \sqrt{3 p r}, a b c=r=1 \\
4\left(1+\frac{1+p}{2+p+q}\right) \leq 4\left(1+\frac{1+p}{2+p+\sqrt{3 p}}\right) \leq 3+p \\
(t-1)\left(3 t^{3}+6 t^{2}+3 t+2\right) \geq 0
\end{gathered}
 [/INST] Using OP's work, WTS for $ a bc = 1,  \sum  a + 1 - \frac{4}{1+a } \geq 0 $.
Use the substitution $ a = e^A, b = e^B, c = e^C$,
WTS for $ A + B + C = 0, \sum e^A + 1 - \frac{4}{ 1 + e^A } \geq 0 $.
Let $ f(x)  = e^x + 1 - \frac{4}{ 1 + e^x }$, then $f'(x) = e^x + 1 + \frac{ 4 e^ x } { (1 + e^x )^2 } $ and $ f''(x) = e^x + 1  - \frac{ 4e^x ( e^x - 1 )} { ( 1 + e^x ) ^3 } = \frac{ e^{4x} + 4e^{3x} + 2e^{2x} + 8e^x + 1 } { ( 1 + e^x) ^3 } \geq 0.$
Hence, we can apply Jensen's to conclude that
$ f(A) + f(B) + f(C) \geq 3 f( \frac{ A + B + C } { 3 } ) = 3 f(0) = 0 $.

Notes:

*

*This approach of using exponential to convert $abc = 1 $ to $ A+B+C = 0 $ and then apply Jensens is a standard trick.

*It is plausible that Jensen's could have worked directly, but it was too ugly for me to want to try. Have at it.

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