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Let $n\geq 3$, $\frac{1}{2} > c_1 \geq c_2 \geq \dots \geq c_n \geq 0$ real, with $c_1 + \dotsc + c_n \leq 1$. Assume moreover that for $j = 1, \dotsc, n-2$ $$ c_j < c_{j+1} + \dotsb + c_n $$ and define $$ c_j^{*} = 1 - \frac{c_j}{c_{j+1} + \dotsb + c_n}. $$

Then (claim) $$ c_1 + c_1^{*}c_2 + c_1^{*}c_2^{*}c_3 + \dotsb + c_1^{*}\dotsm c_{n-2}^{*}c_{n-1} < \frac{1}{2}. $$

Any suggestion/proof/counterexample is welcome!

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By induction. I omit the base case ($n = 3$): I checked it and it works.

Suppose the hypothesis is true for $n-1$.

Let $x := c_2 + \dotsc + c_n$. We know that $c_2 < \frac{x}{2}$. By the induction hypothesis applied to the $(n-1)$-tuple $\bigl(\frac{c_2}{x},\dotsc ,\frac{c_n}{x}\bigr)$ we obtain

$$c_2 + c_2^{\ast} c_3 + \dotsc + c_2^{\ast}\dotsc c_{n-2}^{\ast} c_{n-1} < \frac{x}{2}.$$

Then

$$\begin{align}&\quad c_1+c_1^*c_2+\dotsc+c_1^*\dotsc c_{n-2}^*c_{n-1} \\ &=c_1+c_1^*\bigg(c_2+c_2^*c_3+\dotsc+c_2^*\dotsc c_{n-2}^*c_{n-1}\bigg) \\ &<c_1+c_1^*\frac{x}{2}\le\frac{1}{2}.\end{align}$$

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    $\begingroup$ When $n=3$, we choose $c_1=c_2=c_3=\frac 1 3$, and then $c_1^*=\frac 1 2$. Thus, $$c_1+c_1^*c_2=\frac 1 3 + \frac 1 6=\frac 1 2$$ $\endgroup$ – Aforest Oct 18 '16 at 14:08
  • $\begingroup$ What does ensure that choosing $c_1 = c_2 = c_3$ gives the maximum value? $\endgroup$ – Hugo Oct 18 '16 at 16:22

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