Solving the problem via MVT.

Question

Here is the problem I am trying to solve:

Suppose $f : [1, \infty) \to \mathbb R$ is differentiable with continuous derivative and

(a) $f(1) = 1$

(b) $x f'(x) \leq \frac{1}{2} f(x)$

Show that $f(x) \leq x^{1/2}$ for all $x \geq 1$.

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My trial

I assumed that $f(x) = x^m$ (which satisfies the first assumption) and I used the second assumption together with the derivative of the defined $f(x)$ above to conclude that $m = \frac12$. But I am wondering if we can solve the problem via MVT? Could someone help me in answering this question please?

Answer 1

$(\frac {f(x)} {\sqrt x})'=\frac {\sqrt x f'(x)-\frac 1 2 f(x)x^{-1/2}} {x} \leq 0$ so $\frac {f(x)} {\sqrt x}$ is decreasing. Hence $\frac {f(x)} {\sqrt x} \leq \frac {f(1)} 1=1$. You can apply MVT to $\frac {f(x)} {\sqrt x}-\frac {f(1)} 1$ if you wish, but that looks unnecessary.