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.
 A: let $p=a+b+c,q=ab+bc+ca,r=abc$
Now it is easy to prove the following as $r=1$

*

*$pq\ge 9r=9$

*$p^2\ge 3q$
Now we have to prove $$4(3+2p+q)\le (p+3)(2+p+q)$$ or $$p^2+pq\ge 6+3p+q$$ $$p^2+9\ge 6+3p+q$$ using $q\le p^2/3$ $$\frac{2p^2}{3}-3p+3\ge 0$$ which is true as $p\ge 3$
A: 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.
A: By Schur's inequality and AM-GM $$\begin{align*}x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(x+z)&\\ \ge 2( {(xy)}^{3/2}+{(yz)}^{3/2}+{(zx)}^{3/2})\end{align*}\tag S$$
Now as $$\frac{1}{1+x}\le \frac{1}{2\sqrt{x}}$$ we have to prove $$a+b+c+3\sqrt[3]{abc}\ge 4\left(\frac{1}{2\sqrt{a}}+\frac{1}{2\sqrt{b}}+\frac{1}{2\sqrt{c}}\right)$$ or it suffices to prove $$a+b+c+3\sqrt[3]{abc}\ge 2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}$$ which is just a direct consequence of inequality (S) and so we are done!
Note here $\sqrt[3]{abc}=\sqrt{abc}=1$ is used to normalize inequality
