2

Prove the inequality $ \int_{0}^{1} x \sqrt{1+ \{f'(x)\}^2 } dx \le {{1} \over {\sqrt{2}}} $, where $f(0)=1, f(1)=0$ and $f'(x)<0, f''(x) \ge 0 $ on the interval $(0, 1)$.

My attempt: Put $ \int_{0}^{x} \sqrt{1+ \{f'(t)\}^2} dt = l(x) $.
Then $ \int_{0}^{1} x \sqrt{1+ \{f'(x)\}^2 } dx = l(1) - \int_{0}^{1} l(x)dx = \int_{0}^{1} \int_{x}^{1} \sqrt{1+ \{f'(t)\}^2} dt \space dx $.

But I couldn't make any progress after this.

user896690
  • 21
  • $f'$ cannot be bounded. e.g. $f(x)=1-\sqrt{1-(x-1)^2}$ – user896690 Mar 07 '21 at 09:08
  • $f'(1) \ge -1$ can be obtained from $f(0)=1$ and $f(1)=0$. Suppose $f'(x) \le f'(1) < -1$, then $ \int_{0}^{1}{f'(x)dx} \le \int_{0}^{1}{f'(1)dx} < -1$, which yields contradiction. And one more, $\sqrt{1+f'(x)^2} \ge \sqrt{1+f'(1)^2}$ since $f'(x) \le f'(1) < 0$. – user896690 Mar 07 '21 at 10:10
  • Indeed, what you wrote works (for your first proof). But I've made a mistake (I forgot the condition $f'<0$, sorry). So, in fact, since $f'$ is increasing and negative, $\sqrt{1+f(x)^2}\leq \sqrt{1+f'(0)^2}$. Therefore, $$\int_0^1x\sqrt{1+f'(x)^2}dx\leq \int_0^1 x\sqrt{1+f'(0)^2}dx=\frac{\sqrt{1+f'(0)^2}}{2}.$$ Now, if you have that $f'(0)\geq -1$, then you can conclude (but unfortunately, what you did in your previous comment won't work). – Surb Mar 07 '21 at 10:27
  • $f'(0)>-1 \Rightarrow \int_{0}^{1}{f'(x)dx} \ge \int_{0}^{1}{f'(0)dx} >-1$ so contradiction again. This inequality seems not to be proven in this way... – user896690 Mar 07 '21 at 10:56
  • What I meant to say was that the case $f'(0)>-1$ is impossible, because the equality in $f(1)-f(0) \ge -1$ holds only when $f'(0)=-1$. – user896690 Mar 07 '21 at 11:13
  • So, this method works only if $f'(0)=-1$... – Surb Mar 07 '21 at 11:19
  • 2
    Does this answer your question? How prove this integral inequality $\int_{0}^{1}x \sqrt{1+{f'(x)}^2}dx\le\ \frac{1}{\sqrt{2}}$ – cineel Jan 02 '22 at 13:54

0 Answers0