Exercise 4.7 Kesavan's book Topics in Functional Analysis and Applications Below I present Exercise 4.7 from Kesavan's book Topics in Fucntional Analysis and Applications chapter 4.
I could solve item (a) but I don't know how to get (b) from (a). I would like some hint or reference. 
(a) Let $V$ be a Banach space and $A:D(A)\subset V\to V$ the infinitesimal gererator of a $C_0$-semigroup {$S(t)$} on $V$. If $u\in D(A^2)$, show that 
$$S(t)u-u=tAu+\int_0^t(t-\tau)S(\tau)A^2u\, d\tau.$$
(b) Deduce that for $u\in D(A^2)$,
$$\|Au\|^2\leq 4M^2\|A^2u\|\,\|u\|,$$
if $\|S(t)\|\leq M,$ for all $t.$
Solution of item (a) 
In order to solve (a) I defined $f(t)=S(t)u$. Since $u\in D(A^2)$ we have that $f\in C^2([0,+\infty);V)$. Also $f'(t)=S(t)Au$  and $f''(t)=S(t)A^2u$. Using Taylor Formula with integral remainder we have the result for item (a). 
 A: Finally I found the answer. After a lot of search, I found out, in this book A short course on operator semigroups page 45, that this inequality is Know as Landau–Kolmogorov inequality. So, it was a question of time until to find the proof in the Lemma 2.8 of the classical book Pazy. Semigroups of linear operators and applications to partial differential equations..
For the sake of convenience, I will reproduce the proof here.
From (a) we have that
$$Au = \frac{S(t)u-u}{t}+\frac{1}{t}\int_0^t(t-\tau)S(\tau)A^2u\, d\tau,\ \forall t\in (0,+\infty)$$
Hence, for all $t\in (0,+\infty)$, we have that
\begin{align}
\|Au\|& \leq  \frac{1}{t} \big(\|S(t)u\|+\|u\|\big)+\frac{1}{t}\int_0^t(t-\tau)\|S(\tau)A^2u\|\, d\tau\\
& \leq  \frac{1}{t} \big(M\|u\|+\|u\|\big)+\frac{M}{t}\|A^2u\|\int_0^t(t-\tau)\, d\tau\\
& \leq \frac{1}{t}(M+1)\|u\|+\frac{Mt}{2}\|A^2u\|\\
& \leq \frac{2M}{t}\|u\|+\frac{Mt}{2}\|A^2u\|.\ \ \ \ \  \text{($M\geq 1$, since $S(0)=I$.) }
\end{align}
Now, note that the function $f(t)=\frac{a}{t}+bt$ has a minimum $2\sqrt{ab}$ at $t=\sqrt{\frac{a}{b}}$. Hence, taking $a=2M\|u\|$ and $b=\frac{M}{2}\|A^2u\|$ we have that
$$\|Au\|\leq 2\sqrt{M^2\|A^2u\|\|u\|},$$
whence the result follows.  
It is worth to mention that, if we don't use $M\geq 1$, we will actually prove a the following better inequality
$$\|Au\|^2\leq 2(M+1)M\|A^2u\|\|u\|.$$
This proof is found in  H. Kraljević, S. Kurepa, Semigroups on Banach spaces, Theorem 3.
