I'm trying to show this problem.
Show that $$ f(x):=2(2-a)(3-a)x^{2}-5(3-a)x+6-9(1-x)^{3-a} $$ is positive on $\{x\in\mathbb{R} \mid \frac{2}{3}<x\le1\}$ for all $a\in(0,1)$.
I have tried to remove the fractional order of the last term but I could not show well. More precisely, I estimated as follows: $$ f(x)>2(2-a)(3-a)x^{2}-5(3-a)x+6-9(1-x)^{2}, $$ but the right-hand side becomes negative near $x=\frac{2}{3}$ for $a\approx0.6〜0.7$. Although I consider the above problem since I want to show an inequality $$ 2(2-a)(3-a)x^{2}-5(3-a)x+6>9(1-x)^{3-a}\quad\text{on $\{x \mid 2/3<x\le1\}$}, $$ please tell me if you have another approach.
I'm glad if you give a hint or solution.
Thank you in advance.
hint: let $$G(x)=2(2-a)(3-a)x^2-5(3-a)x+6-9(1-x)^{3-a}$$ then $$G'(x)=4(2-a)(3-a)x-5(3-a)+9(3-a)(1-x)^{2-a}=(3-a)H(x)$$ where $$H(x)=4(2-a)x-5+9(1-x)^{2-a}$$ then $$H'(x)=4(2-a)-9(2-a)(1-x)^{1-a}=(2-a)Q(x)$$ where $$Q(x)=4-9(1-x)^{1-a}$$ you can determine the $Q(x)$ sign?