<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] Remarks: Here is a trick for the inequality
of the form $f(a) + f(b) + f(c) \ge 0$ under the constraints $a, b, c > 0$ and $abc = 1$.
Let $F(x)  = f(x) + m \ln x$. If we can find
an appropriate $m$ such that $F(x)\ge 0$ for all $x > 0$, then we have $F(a) + F(b) + F(c)
= f(a) + f(b) + f(c) + m\ln (abc)  \ge 0$
and we are done.

We apply the trick for our problem.
Let $F(x) = x + 1 - \frac{4}{1 + x} - 2\ln x$.
We have $F'(x) = \frac{(x - 1)(x^2 + x + 2)}{x(x + 1)^2}$.
Thus, $F'(1) = 0$, and $F'(x) < 0$ on $(0, 1)$, and $F'(x) > 0$ on $(1, \infty)$.
Thus, $F(x) \ge F(1) = 0$ for all $x > 0$.
Thus, we have $F(a) + F(b) + F(c) \ge 0$
which results in $4\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right) \leq 3+(a+b+c)$.
We are done.

How to determine the coefficient $m$:
Let $F(x) = x + 1 - \frac{4}{1 + x} + m\ln x$.
We have $F'(x) = 1 + \frac{4}{(1 + x)^2} + \frac{m}{x}$. Let $F'(1) = 0$ and we have $m = -2$.
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