Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f$ be a positive-valued,concave function on $[0,1]$,Prove that $$6\left(\int_{0}^{1}f(x)dx\right)^2\le 1+ 8\int_{0}^{1}f^3(x)dx$$

Let $$A=\int_{0}^{1}f^3(x)dx,B=\left(\int_{0}^{1}f(x)dx\right)^2$$ $$\Longleftrightarrow 6B\le 1+8A$$ let $$F(x)=\int_{0}^{x}f(t)dt$$ $$F(x)=x\int_{0}^{1}f[ux+(1-u)\cdot 0]du\ge x\int_{0}^{1}[uf(x)+(1-u)du=\dfrac{xf(x)}{2}+\dfrac{x}{2}$$

Maybe this problem folowing can use Cauchy-Schwarz inequality to solve it,Thank you

share|cite|improve this question
I always have this doubt... concave from which side? – evil999man Apr 21 '14 at 16:58
Have you tried playing around with Jenson's inequality? – Alex R. Apr 21 '14 at 18:17
Don't have much time to work on this now, some thoughts:if $f(0) = 0$ then by concavity $f$ is also subadditive, i.e. $f(x+y) \leq f(x) + f(y)$. Moreover, Jensen's (or Cauchy-Schwarz for $\cdot \rightarrow \cdot^2$) implies $\left(\int_0^1 f\right)^2 \leq \int_0^1 f^2$, whereas concavity gives $\int_0^1 f(x) \leq f(x^2/2)$. Finally, am I right to say $f'$ exists a.e.? – snarski Apr 22 '14 at 0:47
To get the $3$ to come out on the $f^3$, perhaps apply Holder to $\|f\|_2 \leq 1\cdot \|f\|_{3/2}$? – snarski Apr 22 '14 at 0:54
@snarski,maybe can use Holder inequality,But I think this not – math110 Apr 22 '14 at 0:56

Let $c=\int_0^1 f(x)\,dx$ and $g=f/c$, so $\int_0^1 g(x)\,dx=1$. Then by Holder's inequality, $$ 1\le \left(\int_0^1 g(x)^3\,dx\right)^{1/3}\left(\int_0^1 1^{3/2}\,dx\right)^{2/3} . $$ Therefore $\int_0^1 f(x)^3\,dx=c^3\int_0^1g(x)^3\,dx\ge c^3$, and $$ 8\int_0^1 f(x)^3\,dx + 1-6\left(\int_0^1 f(x)\,dx\right)^2\ge 8c^3+1-6c^2 =: h(c). $$ For $c>0$ the right-hand side is minimized when $0=h'(c)=24c^2-12c$, meaning $c=1/2$ (noting $h'(c)<0$ for $c<1/2$ and $h'(c)>0$ for $c>1/2$). Thus $$h(1/2)=8(1/2)^3+1-6(1/2)^2=1/2\le h(c)$$ for all $c>0$. Actually, then it follows
$$ 6\left(\int_0^1 f(x)\,dx\right)^2 \le \frac12 + 8\int_0^1f(x)^3\,dx. $$ Concavity of $f$ is not needed.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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