Proving $ \frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \leq 1 $ for $0Let $a,b,c$ be positive real numbers between $0$ and $1$ ,i.e., they lie in the closed interval $[0,1]$. Prove that :
$$ \frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \leq 1 $$
 A: Without loss of generality
let $0\le a\le b\le c\le 1$,then
\begin{align*}
&\dfrac{a}{c+b+1}+\dfrac{b}{a+c+1}+\dfrac{c}{a+b+1}+(1-a)(1-b)(1-c)\\
&\le\dfrac{a}{a+b+1}+\dfrac{b}{a+b+1}+\dfrac{c}{a+b+1}+(1-a)(1-b)(1-c)\\
&=\dfrac{a+b+c}{a+b+1}+\dfrac{(a+b+1)(1-a)(1-b)(1-c)}{a+b+1}\\
&\le\dfrac{a+b+c}{a+b+1}+\dfrac{(a+1)(b+1)(1-a)(1-b)(1-c)}{a+b+1}\\
&=\dfrac{a+b+c}{a+b+1}+\dfrac{(1-a^2)(1-b^2)(1-c)}{a+b+1}\\
&\le\dfrac{a+b+c}{a+b+1}+\dfrac{1-c}{a+b+1}=1
\end{align*}
A: Let's prove  the inequality
$$\sum_{i=1}^n \frac{a_i}{\sum_{j\ne i} a_j + 1} + (1-a_1) \cdots (1-a_n) \le 1$$
for $a_i \in [0,1]$.
For fixed $a_j$, $j\ne i$ the function is strictly convex in $a_i$. We conclude that for every $a=(a_i)$ we have
$$f(a) \le \max_{v \in \{0,1\}^n} f(v)$$
Now consider $v \in\{0,1\}^n$, $k$ the number of components $=1$. If $k=0$, $v=(0,0,\ldots, 0)$ we have
$f(v) = 0+1 = 1$. If $k>0$, we have $k$ terms in the sum that equal $\frac{1}{k}$, while the product equals $0$. We conclude $f(v) = 1$ for all $v\in \{0,1\}^n$. The conclusion follows.
A: Let $f(a,b,c)=\sum\limits_{cyc}\frac{a}{b+c+1}+\prod\limits_{cyc}(1-a)$.
Hence, $\frac{\partial^2f}{\partial a^2}=\frac{2b}{(a+c+1)^3}+\frac{2c}{(a+b+1)^3}\geq0$,
which says that $f$ is a convex function of $a$, of $b$ and of $c$.
Thus, $$\max_{\{a,b,c\}\subset[0,1]}f=\max_{\{a,b,c\}\subset\{0,1\}}f=f(1,1,1)=1.$$
Done!
