How prove this mathematical analysis by zorich from page 233 Let $f$ be twice differentiable on an interval $I$,Let

$$M_{0}=\sup_{x\in I}{|f(x)|},M_{1}=\sup_{x\in I}{|f'(x)|},M_{2}=\sup_{x\in I}{|f''(x)|}$$

show that

(a):$$M_{1}\le 2\sqrt{M_{0}M_{2}}$$
  if the length of $I$ is not less than $2\sqrt{\dfrac{M_{0}}{M_{2}}}$

(b):the numbers $2$ and $\sqrt{2}$ (in part a) cannot be replaced by smaller numbers.

My try:for part $(a)$ I can prove if the length of $I$ is not less than $4\sqrt{\dfrac{M_{0}}{M_{2}}}$

my proof:

$$f(c)-f(x)=f'(x)(c-x)+\dfrac{f''(x+\theta_{1}(c-x))}{2}(c-x)^2$$
  let $c-x=h$
  then we have
  $$f'(x)=\dfrac{f(c)-f(x)}{h}-\dfrac{f''(x+\theta_{1}h)}{2}h$$
  Thus
  $$f'(x)\le\dfrac{2M_{0}}{h}+\dfrac{1}{2}M_{2}h$$
  Now taking $h=2\sqrt{\dfrac{M_{0}}{M_{2}}}$, which in turn implies that
  $$M_{1}\le2\sqrt{M_{0}M_{2}}$$
  and $$c=x+h=2\sqrt{\dfrac{M_{0}}{M_{2}}}+x>2\sqrt{\dfrac{M_{0}}{M_{2}}}$$
  so  the length of $I$ is $2c$ and  not less than $4\sqrt{\dfrac{M_{0}}{M_{2}}}$

so Now  How part(a)?
and for part(b) How prove it? and How take example for the best numbers$\sqrt{2}$?
By the way
This problem is from Mathematical Analysis I(Zorich )Page 233 problem 9.
Thank you evryone
 A: (a) We should assume that $M_2>0$. Denote the length of $I$ by $|I|$ and denote 
$$l=2\sqrt{\frac{M_0}{M_2}}.$$ 
If $|I|\ge l$, then for any $x\in I$, there exist $h,k\ge 0$, such that $x+h,x-k\in I$ and $h+k=l$. As shown in your argument, using Taylor's expansion, we have:
$$f(x+h)=f(x)+f'(x)h+\frac{f''(\xi)}{2}h^2\tag{1}$$
and
$$f(x-k)=f(x)-f'(x)k+\frac{f''(\zeta)}{2}k^2\tag{2},$$
where $x\le\xi\le x+h$ and $x-k\le\zeta\le x$.
Noting that $h^2+k^2\le l^2$, $(1)-(2)$ yields
$$|f'(x)|\cdot l\le |f(x+h)|+|f(x-k)|+\frac{1}{2}\left(|f''(\xi)|h^2+|f''(\zeta)|k^2\right)\le 2M_0+\frac{M_2}{2}l^2,$$
and the conclusion follows. 

(b) The best constant in (a) should be $2$. That is to say, 
(i) there exists $f:I\to\Bbb R$, such that $|I|=l=2\sqrt{\frac{M_0}{M_2}}$ and $M_1=2\sqrt{M_0M_2}$; 
(ii) for every $0<\lambda<1$, there exists $f:I\to\Bbb R$, such that $I\ge \lambda l$ and $M_1>2\sqrt{M_0M_2}$. 
Example: For any $a\in [0,1)$, let 
$$f(x)=2x^2-1-a^2,\quad I=[a,1].$$ Then $|I|=1-a$,  $M_0=1-a^2$, $M_1=4$, $M_2=4$, $l=\sqrt{1-a^2}$ and $2\sqrt{M_0M_2}=4\sqrt{1-a^2}$. 
(i) When $a=0$, $|I|=l=1$, and  $M_1=2\sqrt{M_0M_2}=4$.
(ii) Given $\lambda\in (0,1)$, let $a=\frac{1-\lambda^2}{1+\lambda^2}\in (0,1)$. For this $a$, $|I|=\lambda l$ and $M_1=4>2\sqrt{M_0M_2}$.
A: Let's assume that $f\geq0$. Then
$$
f(x+t)=f(x)+f'(x)t+f''(\xi)t^2/2\geq0.
$$
In particular, the points where the equation (for $t$) vanishes are
$$
t=\frac{-f'\pm\sqrt{(f')^2-4ff''}}{2f}.
$$
From here we have
$$
(f')^2\leq 4ff''\rightarrow f'\leq 2\sqrt{ff''}.
$$
Here the length of the interval plays no role. I think that this inequality is called Landau inequality. I hope this helps.
