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

Question

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 $x \in (0,1)$?

Answer 1

One may see that for the initial inequality to hold true it is sufficient to prove that $$ 1-\frac{(2-x) (3-x) \left(36-84 x+89 x^2-50 x^3+13 x^4\right)}{16 x(1-x)^2 }<0,\quad x \in (0,1), \tag1 $$ 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 $(2)$.

Answer 2

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