Showing that $\sqrt{f(x)}\ge \sqrt{f(1)}+\frac{1}{2}(x-1) \space \forall \space x\ge 1$ if $f^{'}(t)\ge \sqrt{f(t)}$ for all $t$

Question

Also Given that:

Suppose $f$ is a function such that $f(x)>0$ and $f^{'}(x)$ is continuous at every real number $x$

Now we can write $f(t)\ge \sqrt{f(0)}+\int_{0}^{t} \sqrt{f(t)} dt= \sqrt{f(0)}+\sqrt{f(1)}+\int_{1}^{t} \sqrt{f(t)} dt$

Now since $f(0)=0$

Thus $f(t)\ge \sqrt{f(1)}+\int_{1}^{t} \sqrt{f(t)} dt$

I am quite certain that this is wrong. Give me some hints to solve this please.

Solution :

In line with what @GAVD and @grand_chat suggested

"Hint: The Mean Value Theorem states $$ g(x) - g(1) = g'(t)(x-1) $$ for some $t$ between $1$ and $x$. The obvious choice for $g$ is $g(x):=\sqrt{f(x)}$."

We see $g^{'}(t)=\dfrac{f^{'}(t)}{2\sqrt{f(t)}}\ge \dfrac{1}{2}$

$\therefore \sqrt{f(x)}-\sqrt{f(1)}\ge \dfrac{1}{2}(x-1)$

To tell you honestly I also thought of it in lines of the mean value theorem but was confused with that

$g'(t)\ge \dfrac{1}{2}$

Answer 1

Hint: The Mean Value Theorem states $$ g(x) - g(1) = g'(t)(x-1) $$ for some $t$ between $1$ and $x$. The obvious choice for $g$ is $g(x):=\sqrt{f(x)}$.

Answer 2

Hint: Observe \begin{align} \frac{f'(t)}{2\sqrt{f(t)}}= \frac{d}{dt}\sqrt{f(t)}\geq \frac{1}{2}. \end{align}

Answer 3

Let $g(x) = \sqrt{f(x)} - \frac{x}{2}$. One then has $g'(x) = \frac{f'(x)}{2\sqrt{f(x)}} - \frac{1}{2} > 0$. Then you have $g(x) \geq g(1)$, $\forall x \geq 1$.