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:
$$\frac{1}{1-x} - \frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)(2x(1-x))^2} < 1,$$
for all $x \in (0,1)$?
 A: 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 $(2)$.
A: $$\frac{1}{1-x} - \frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)(2x(1-x))^2} < 1,$$
$$\frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)(2x(1-x))^2} > \frac{1}{1-x}-1,$$
$$\frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)(2x(1-x))^2} > \frac{x}{1-x}.$$
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.
