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Can anyone suggest any hints to prove the following inequality:

$$\frac{1}{1-x} - \frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)(2x(1-x))^2} < 1,$$

for all $x \in (0,1)$?

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One may see that for the initial inequality to hold true it is sufficient to prove that $$ 1-\frac{(2-x) (3-x) \left(36-84 x+89 x^2-50 x^3+13 x^4\right)}{16 x(1-x)^2 }<0,\quad x \in (0,1), \tag1 $$ then setting $$ \begin{align} &f(x)=16 x(1-x)^2-(2-x) (3-x) \left(36-84 x+89 x^2-50 x^3+13 x^4\right) \end{align} $$ one gets$$ \begin{align} &f'(x)=700-2044 x+2535 x^2-1668 x^3+575 x^4-78 x^5 \in [20,700] \tag2 \\\\&f'(x)>0\implies f \nearrow, \quad x \in (0,1), \quad f(0)=-216, \quad f(1)=-8, \end{align} $$ giving $$ f(x)<0, \quad x \in (0,1), \tag3$$ which yields $(1)$ then yielding the initial inequality.

Remark. By setting $t=1-x$ in $f'(x)$ above, one gets a polynomial with positive coefficients:

$$ 20+68 t+201 t^2+148 t^3+185 t^4+78 t^5 >0,\quad t \in (0,1), $$

proving $(2)$.

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  • $\begingroup$ I tried this way, too but I was stuck at giving a proper argument (why it holds $f'(x)\in [20,700]$) without using a calculation software. Do you know a better way? $\endgroup$ – Nogard Jul 20 '16 at 9:04
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    $\begingroup$ @Nogard Yes I do. Edited. Thanks. $\endgroup$ – Olivier Oloa Jul 20 '16 at 9:34
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    $\begingroup$ That's very elegant. Thanks! $\endgroup$ – Nogard Jul 20 '16 at 9:54
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$$\frac{1}{1-x} - \frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)(2x(1-x))^2} < 1,$$ $$\frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)(2x(1-x))^2} > \frac{1}{1-x}-1,$$ $$\frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)(2x(1-x))^2} > \frac{x}{1-x}.$$ For $x\in(0,1)$ we can prove that $$\frac{(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(2x(1-x))^2} > 1,$$ or $$(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36) - 4(2x(1-x))^2 >0.$$ Taking in account that for $x\in(0,1)$ $$x(1-x)\leq \dfrac14,\quad 3-x>2,\quad 2-x>1,$$ $$13x^4 - 50x^3 + 89x^2 - 84x + 36 = 0.5x^4 + 12.5(x-1)^4 +\dfrac1{14}(14x-17)^2+\dfrac{20}7\geq \dfrac{20}7,$$ easy to prove the required inequality.

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Hint #1: Assume to the contrary that $$\frac{1}{1-x} - \frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)\left(2x(1-x)\right)^2} \geq 1$$ and then derive a contradiction to $0 < x < 1$.

Hint #2: (Note: This edit is in response to Joshua Lochner.) So the proof by contradiction actually goes like this. Assume to the contrary that there exists an $x \in \mathbb{R}$ such that $$\frac{1}{1-x} - \frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)\left(2x(1-x)\right)^2} \geq 1.$$ By this WolframAlpha computation, this implies that $$\left\{x \geq 3.9483\right\} \lor \left\{1 < x \leq 1.41066\right\} \lor \left\{x < 0\right\},$$ which contradicts the assumption $0 < x < 1$. Hence, we conclude that $$\forall x \in (0, 1), \hspace{0.5in} \frac{1}{1-x} - \frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)\left(2x(1-x)\right)^2} < 1$$

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    $\begingroup$ Wouldn't that just mean testing each Real number? Which cannot be done. It would be the same as testing x = 0.5, getting a contradiction and assuming this is true for all values $0 < x < 1$, which cannot be done, as there may be one example that does work. $\endgroup$ – Joshua Lochner Jul 20 '16 at 9:10
  • $\begingroup$ @JoshuaLochner, I refer you to mathworld.wolfram.com/ProofbyContradiction.html. $\endgroup$ – Arnie Bebita-Dris Jul 20 '16 at 9:15
  • $\begingroup$ But finding one contradiction for $0<x<1$ will not automatically mean this is true for all values $0<x<1$ $\endgroup$ – Joshua Lochner Jul 20 '16 at 9:19
  • $\begingroup$ @JoshuaLochner, $x$ is taken to be an arbitrary real number in the interval $(0, 1)$. $\endgroup$ – Arnie Bebita-Dris Jul 20 '16 at 9:59
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    $\begingroup$ Why assume anything if you are going to calculate it? This is not a proof at all. $\endgroup$ – Ennar Jul 20 '16 at 10:15

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