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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.

River Li
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    Welcome to Math SE. FYI, using Approach0, I found the AoPS thread Inequality, with message #$9$ from MariusStanean being of particular interest to you. – John Omielan Mar 12 '22 at 07:23
  • @John Omielan: What’s approacho? By the way I know AoPS. –  Mar 12 '22 at 07:35
  • Approach0 is a search engine allowing you to search (in particular for mathematical expressions) this site and/or AoPS (Art of Problem Solving) (I don't think it supports any other sites (e.g., I've not seen any other sites being shown in search results where no site restrictions are used), but I'm not sure). You can get more details about Approach0 in this meta answer. – John Omielan Mar 12 '22 at 07:39

1 Answers1

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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.

River Li
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  • I cannot find simpler what do you think about my second answer ?(+1) – Miss and Mister cassoulet char Mar 14 '22 at 16:38
  • @ErikSatie It seems incorrect. If $(a,b,c,d) = (1/3, 1/4, 1/10, 1/10)$, there is no $p, u, k > 0$ such that $\frac{pa\left(a+u\right)}{a+k}=b,\frac{pb\left(b+u\right)}{b+k}=c,\frac{pc\left(c+u\right)}{c+k}=d$ (you get $k = -9/41, p = 36/41, u = -17/72$). – River Li Mar 14 '22 at 23:09