Prove that $\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}\le2$ for $0 \le a, b, c, d \le 1$ 
Let $0\le a,b,c,d\le 1$, prove that $$\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}\le2.$$

I’ve solved it, but I want a solution without derivatives. My solution is posted below.
The part of the expression concerning $a$ is $\frac{a}{1+b}+ \frac{d}{1+a}$, the second derivative of which is $0\le\frac{2d}{1+a^3}$. This means that to reach its maximum value, there must be $a=0$ or $a=1$. Similarly, $b,c,d$ must be either $1$ or $0$. Plug in the values to find it does not exceed 2. Done.

Note that, for all $x\in [0, 1]$,
$$1 - x/2 - \frac{1}{1 + x}
= \frac{x(1 - x)}{2(1 + x)}\ge 0.$$
We have
\begin{align*}
 &\frac{a}{1 + b} + \frac{b}{1 + c}
 + \frac{c}{1 + d} + \frac{d}{1 + a}\\
 \le\, & a (1 - b/2) + b(1 - c/2) + c(1 - d/2) + d(1 - a/2)\\
 =\, & a + b + c + d - \frac12(ab + bc + cd + da)\\
 =\,& a + b + c + d - \frac12(a + c)(b + d)\\
 =\,& \frac12 (a + c) (2 - b - d) + b + d\\
 \le\, &\frac12\cdot 2\cdot (2 - b - d) + b + d\\
 =\,& 2
\end{align*}
where we have used $a + c \le 2$ and $b + d \le 2$.
We are done.
