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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

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I always have this doubt... concave from which side? –  Awesome Apr 21 at 16:58
    
Have you tried playing around with Jenson's inequality? –  Alex R. Apr 21 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 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 at 0:54
    
@snarski,maybe can use Holder inequality,But I think this not –  math110 Apr 22 at 0:56

1 Answer 1

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.

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