# Help with an inequality with integrals [duplicate]

Possible Duplicate:
Proving Integral Inequality

Let $f$ be defined on $[0,1]$ with $f(0)=0$ and $0 < f'(x) \leq 1$. Prove that $$\int_{0}^{1} f^3(x)\ dx \leq \bigg[ \int_{0}^{1} f(x)\ dx\bigg]^2$$

Now I have been able to show that $0 < f(x) \leq x$ from this it is simple to show that $f(x) \leq 1$ and hence $f^3(x) \leq f(x)$ thus $\int_{0}^{1} f^3(x)\ dx \leq \int_{0}^{1} f(x)\ dx$.

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 Are you also given that $f(0) = 0$? Try to look at the function $g(t) = \left(\int_0^t f(x)dx\right)^2 - \int_0^t f^3(x)dx$ and study its derivatives. – Sanchez Jan 18 at 2:50 If $f(0) = 0$ then as you said, $0 \leq f(x) \leq x$, whence $\int f dx \leq 1/2$, and squaring we have $(\int f dx )^2 \leq 1/4$. From what you have so far, this completes the inequality. – A Blumenthal Jan 18 at 2:52 How could $(\int f\ dx)^2 \leq 1/4$ complete the inequality? – Eric Jan 18 at 3:36

## marked as duplicate by user17762, Henry T. Horton, Douglas S. Stones, EricJan 18 at 5:22

Let $f(x)=x+c$ where $c$ is some constant. Then $f^{'}(x)=1$, and so the condition is satisfied. Evaluating the left and right hand side, we find that $$\int_0^1 f(x)^3 dx =c^3+\frac{3}{2}c^2+c+\frac{1}{4},$$ and $$\left(\int_0^1 f(x)\right)^2 =c^2+c+\frac{1}{4}.$$ The inequality $$c^3+\frac{3}{2}c^2+c+\frac{1}{4}\leq c^2+c+\frac{1}{4}$$ is evidently false for large $c$. For example, $c=1$ fails. That is the function $f(x)=x+1$ satisfies $0< f^{'}(x)\leq 1.$
 I am so sorry I didn't realize that I forgot to put $f(0) = 0$. As someone above has pointed out this is a duplicate of another problem. Thank you so much for the effort though. I will accept your answer. – Eric Jan 18 at 3:40