Prove the Inequality $\frac{1}{1-x}-\frac{x(3-x)(2-x)(13x^4-50x^3+89x^2-84x+36)}{4(1-x)(2x(1-x))^2}<1$ Can anyone suggest any hints to prove the following inequality:

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

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


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


proving $(2)$.
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 $0 &lt; x &lt; 1$.

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
$$\left\{x \geq 3.9483\right\} \lor \left\{1 &lt; x \leq 1.41066\right\} \lor \left\{x &lt; 0\right\},$$
which contradicts the assumption $0 &lt; x &lt; 1$.  Hence, we conclude that
$$\forall x \in (0, 1), \hspace{0.5in} \frac{1}{1-x} - \frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)\left(2x(1-x)\right)^2} &lt; 1$$