<s>[INST] Prove the following inequality without using differentiation Let $a,b,c$ be real numbers that satisfy $0\le a,b,c\le 1$. Show that 
$$\frac a{b+c+1} + \frac b{a+c+1}  + \frac c{a+b+1} + (1-a)(1-b)(1-c) \le 1.$$
I don't know where to start. Multiplying everything by the denominators creates extreme mess.
 [/INST] WLOG: $a\le b\le c$, then
\begin{align*}&\dfrac{a}{b+c+1}+\dfrac{b}{c+a+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{(1+a)(1+b)(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*}
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