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I tried to solve this inequality: $$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}+\frac{d}{1-d}+\frac{e}{1-e}\ge\frac{5}{4}$$ with $$a+b+c+d+e=1$$ I am stuck at this. I don't want the full solution, a hint would be enough.

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    $\begingroup$ what is the nature of $a,b,c,d,e?$ $\endgroup$ Commented May 25, 2013 at 12:59
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    $\begingroup$ @labbhattacharjee Good point, the inequality is false for $a,b,c,d,e \notin (0,1)$. $\endgroup$ Commented May 25, 2013 at 13:06
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    $\begingroup$ @ABlumenthal, I think they are positive real numbers so that each $<1$ $\endgroup$ Commented May 25, 2013 at 13:07
  • $\begingroup$ yes, i forgot to say this again. $\endgroup$
    – D180
    Commented May 25, 2013 at 14:47

4 Answers 4

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Hint:

You have $1-a=b+c+d+e$ and $a=1-(b+c+d+e)$

$\dfrac{1-(b+c+d+e)}{b+c+d+e}=\dfrac{1}{b+c+d+e}-1$ and $b+c+d+e=1-a$

Using AM $\ge$ HM

$(\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}+\dfrac{1}{1-d}+\dfrac{1}{1-e}) \ge \dfrac{5^2}{1+1+1+1+1-(a+b+c+d+e)} $

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  • $\begingroup$ I thought about the first hint, but I didn't try to use AM/HM. $\endgroup$
    – D180
    Commented May 25, 2013 at 14:49
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HINT:

$$\frac a{1-a}=-1+\frac1{1-a}$$

Assuming $a,b,c,d,e$ to be positive real numbers, $1-a>0 $

Using AM HM inequality on $1-a$ etc

$$\frac{\sum (1-a)}5\ge \frac5{\sum \frac1{1-a}}\implies \sum \frac1{1-a}\ge\frac{5^2}{5-\sum a}=\frac{25}{5-1}$$

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Hint: with $g(a,b,c,d,e) = a + b + c + d + e$ and $$f(a,b,c,d,e) = \frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}+\frac{d}{1-d}+\frac{e}{1-e}$$ we can consider optimizing $f$ over the set of $(a,b,c,d,e)$ for which $g(a,b,c,d,e) = 1$. The method of lagrange multipliers gives a necessary condition for the existence of a critical point on this set: $\nabla f = \lambda \nabla g$ for some $\lambda \in \mathbb{R}$.

However we know that $\nabla g$ is the column vector $(1,1,1,1,1)$, which means that $$ \frac{1}{(1-x)^2} = \lambda $$ for each $x = a,b,c,d,e$ individually. Notice that this condition is the same for each coordinate.

What can you conclude about the solutions to the lagrange multiplier problem, and how can you connect this information to the inequality you wish to prove?

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  • $\begingroup$ I assumed in this hint that the optimization problem is carried out in the reals. $\endgroup$ Commented May 25, 2013 at 13:04
  • $\begingroup$ You need to show existence of a minimum to use this technique. Since you cannot acheive minimum if one of the components is close to $1$, you can assume you are working over $[0,1-\varepsilon] \times \dots \times [0,1-\varepsilon]$ (intersected with the constraint). The set is compact, the function is continuous, you have a minimum. It will not be on the boundary if you compute what happens then, hence must happen in the interior on which you have a Lagrange condition which characterizes it. $\endgroup$ Commented May 25, 2013 at 13:21
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A proof in one line: $$\sum_{cyc}\left(\frac{a}{1-a}-\frac{1}{4}\right)=\sum_{cyc}\left(\frac{(5a-1)}{4(1-a)}-\frac{5}{16}(5a-1)\right)=\sum_{cyc}\frac{(5a-1)^2}{16(1-a)}\geq0.$$ Done!

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