Prove that $\frac1{1+a}+\frac1{1+b}+\frac1{1+c}\leq\frac14(a+b+c+3)$ if $abc=1$ I found the following exercise in a problem book (with no solutions):

Given $a,b,c>0$ such that $abc=1$ prove that
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\leq\frac{a+b+c+3}{4}$$

I tried AM-GM for the fraction on the LHS but got stuck from there.

Let $f(x) = \frac{1}{1 + x} - \frac{x+1}{4} + \frac{1}{2}\ln x$.
We have $f'(x) = \frac{(1-x)(x^2+x+2)}{4x(1+x)^2}$.
Thus, $f'(x) > 0$ on $(0, 1)$, and $f'(x) < 0$ on $(1, \infty)$.
Thus, $f(x)$ is strictly increasing on $(0, 1)$, and strictly decreasing on $(1, \infty)$.
Also $f(1) = 0$. Thus, $f(x) \le 0$ on $(0, \infty)$.
Thus, $f(a) + f(b) + f(c) \le 0$ which results in $\frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c}
\le \frac{a+b+c+3}{4}$ (using $\ln (abc) = 0$).
We are done.
