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Known: $$0<x<1$$ $$0<k<1$$ $$0<x+k<1$$ $$y\geq1$$ $y$ is a whole number. Prove: $$ \frac{xy}{1-(1-x)^y}\leq \frac{(x+k)y}{1-(1-(x+k))^y}. $$

I ran some sample results and the inequality appears to be true. For example: $((0.005*20)/(1-(1-0.005)^9))<((0.006*20)/(1-(1-0.006)^9 )) $

Basically, both denominators and numerators increase slightly from left to right, which is not obvious whether or not this will increase the right side. It is obvious that denominators on both sides will be between 0 and 1.

Since both denominators have to be positive, I first thought about cross-multiplying but could not countinue to simply further: $$ (1-(1-x)^y) ((x+k)y)\leq (xy) (1-(1-(x+k))^y) $$

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  • $\begingroup$ Let $x=10$. That disproves your inequality $0<x<1$. $\endgroup$ – user10354138 Nov 2 '18 at 2:26
  • $\begingroup$ x cannot be 10. 0<x<1 $\endgroup$ – Chris Nov 2 '18 at 2:27
  • $\begingroup$ ... which is why it disproves your assertion $0<x<1$ in your original post. $\endgroup$ – user10354138 Nov 2 '18 at 2:29
  • $\begingroup$ 0<x<1 is a condition for the inequality. x cannot be 10. @user10354138 $\endgroup$ – Chris Nov 2 '18 at 2:30
  • $\begingroup$ Where does it say so? Your post started with no English words, just a list of 5 inequalities of which the second one is $0<x<1$. $\endgroup$ – user10354138 Nov 2 '18 at 2:31
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$$\frac{xy}{1-(1-x)^y}\leq \frac{(x+k)y}{1-(1-(x+k))^y}$$

Introduce $z=x+k$, $z\in(0,1)$. The inequality becomes:

$$\frac{x}{1-(1-x)^y}\leq \frac{z}{1-(1-z)^y}$$

Make another substitution:

$$u=1-x\\v=1-z$$

Obviously, $x<z\implies u>v$. The inequality now becomes:

$${1-u \over 1-u^y}\le{1-v \over 1-v^y}$$

$${1-u \over {(1-u)(1+u+u^2+...+u^{y-1})}}\le {1-v \over {(1-v)(1+v+v^2+...+v^{y-1})}}$$

$${1 \over {1+u+u^2+...+u^{y-1}}}\le {1 \over {1+v+v^2+...+v^{y-1}}}$$

$$1+u+u^2+...+u^{y-1}\ge 1+v+v^2+...+v^{y-1}$$

...which is obviously true for $u>v$ and $y\ge1$.

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  • $\begingroup$ Your proof is right for the question as stated, when $y$ is a whole number. But in fact the inequality is true regardless of whether $y\in\mathbb N$, it's true for all real $y\geq 1$, as can be shown with the calculus approach. Still well done though! $\endgroup$ – YiFan Nov 4 '18 at 0:07
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Hint: We can reframe the problem to showing that the function $$f_a(x):=\frac{ax}{1-(1-x)^a}$$ is either monotonely increasing or constant on $[0,1]$, which is the same as showing that $f'_a(x)\geq0$. Differentiating: $$\begin{split} f'_a(x) = \frac{a(1-(1-x)^a)-a^2x(1-x)^{a-1}}{(1-(1-x)^a)^2}. \end{split}$$ The denominator is always positive, so we are reduced to showing that whenever $x\in(0,1)$ and $a\geq1$, then $$ax(1-x)^{a-1}\leq 1-(1-x)^a.$$

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  • $\begingroup$ For $x=1$ your inequality is not true. $\endgroup$ – Oldboy Nov 3 '18 at 12:18
  • $\begingroup$ You're right, the inequality was pointing the wrong way before. Thanks for the correction, it is fixed now. $\endgroup$ – YiFan Nov 3 '18 at 12:28
  • $\begingroup$ I think that I have found a much simpler proof, please check :) $\endgroup$ – Oldboy Nov 3 '18 at 13:53

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