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.
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What are your thoughts? What have you tried? – AlexR Jan 13 '15 at 07:01
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@AlexR I added +3 in the LHS but that didn't work, i think this is more complicated than just simplifying the functions in LHS... – user197137 Jan 13 '15 at 07:02
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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*}
math110
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Let $f(a, b, c)$ denote the left hand side of the inequality. Since $$\frac{\partial^2}{\partial a^2}f=\frac{2b}{(a+c+1)^3}+\frac{2c}{(a+b+1)^3}\ge0$$we have that $f$ is convex in each of the three variables; hence, the maximum must occur where $a, b, c \in \{0, 1\}$. Since $f$ is $1$ at each of these $8$ points, the inequality follows.
RE60K
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i've solved it with using it that's an easy one. but just elementary algebra, it is quite difficult – user197137 Jan 13 '15 at 07:24