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Let $(a,b) \in \mathbb{R}^{2}$ such that $a<b$ and $f\in C^2([a,b],\mathbb{R})$ such that $f'\neq 0$ and $f''/f'$ is decreasing.

Prove that :

$$|f(b)-f(a)|\geqslant (b-a) \sqrt{f'(a) f'(b)}$$

My attempt:

By the MVT theorem, we have to prove :

$$ |f'(c)| \geq \sqrt{f'(a)f'(b)}, \quad c\in (a,b) $$

Without loss of generality we can assume that $f'$ is stricly positive.

Thus, by the fact that $\ln(f')$ is concave we get $$ \frac{\ln(f'(c))-\ln(f'(a))}{c-a}\geqslant\frac{\ln(f'(b)-\ln(f'(c))}{b-c} $$ So, $$ \frac{\ln \frac{f'(c)}{f'(a)}}{c-a}\geqslant\frac{\ln \frac{f'(b)}{f'(c)}}{b-c} $$

Notice that I haven't use $\frac{f''(a)}{f'(a)}\geq \frac{f''(b)}{f'(b)}$,

So I think what I am doing it's not enough.

Any ideas?

Thank you in advance for your help,

Seirios
  • 33,157

1 Answers1

4

Abbreviate $f'=g$. Please note that you can not use the MVT, as $$|g(c)|\geq \sqrt{g(a)g(b)},\quad \forall c\in (a,b) $$ is not possible if $g(a)\neq g(b)$. Instead we note that (wlog $g>0$) $$|f(b)-f(a)|=\int_a^b g(x)dx=(b-a)\int_0^1g((1-s)a+bs)ds. $$ As $g$ is log concave, we have that $$g(sb+(1-s)a)\geq g(b)^sg(a)^{1-s}.$$ This gives us $$|f(b)-f(a)|\geq (b-a)\int_0^1g(b)^sg(a)^{1-s}ds =(b-a)\frac{g(b)-g(a)}{ln(g(b))-ln(g(a))}.$$ The inequality $$ \frac{g(b)-g(a)}{ln(g(b))-ln(g(a))} \geq \sqrt{g(a)g(b)} $$ is not too difficult to proof, I will leave it to you. (Please ask if you need help).

Leverkuehn
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