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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As stated, the problem is false. Likely you need another condition which guarantees that we cannot shift the function by a constant. 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 $0< f^{'}(x)\leq 1.$ |
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