# Prove that:$\int_{a}^{b}(2x^{3}-3(a+b)x^{2}+6abx)f'(x)dx\geq (a-b)^{3}f(a)$.

Let $$f:[a,b]\rightarrow[0,\infty)$$ a differentiable function with its derivative continuous and $$f(a)=f(b)$$. Prove that:$$\int_{a}^{b}(2x^{3}-3(a+b)x^{2}+6abx)f'(x)dx\geq (a-b)^{3}f(a)$$. I tried to use Cauchy-Schwarz inequality by refining the expression in the left side but I couldn't finish it. Integration by parts gives a progress but not definitive. EDIT: The integral is, by parts, $$[2( b^3-a^3)-3(a+b)(b^2-a^2)+6ab(b-a)]f(a)-6\int_{a}^{b}(x-a)(x-b)f(x)dx$$=$$(a-b)^3f(a)-6\int_{a}^{b}(x-a)(x-b)f(x)dx$$.

You evaluated the boundary term wrong in that integration by parts; it should be $$[2(b^3-a^3)-3(a+b)(b^2-a^2)+6ab(b-a)]=(b-a)(-b^2+2ab-a^2)=(a-b)^3$$ times $$f(a)$$. So then, the problem becomes showing that $$(a-b)^3f(a)+6\int_a^b(x-a)(b-x)f(x)\,dx \ge (a-b)^3f(a)$$ Since $$f(x)\ge 0$$ for all $$x\in [a,b]$$, $$\int_a^b(x-a)(b-x)f(x)\,dx\ge 0$$ and we have the desired inequality.