1
$\begingroup$

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, $[2( b^3-a^3)-3(a+b)(b^2-a^2)+6ab(b-a)]f(a)-6\int_{a}^{b}(x-a)(x-b)f(x)dx$=$(a-b)^3f(a)-6\int_{a}^{b}(x-a)(x-b)f(x)dx$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.