I'm having some trouble proving the following inequality:
$$\sqrt x \leq \frac{x+1}{2}\ , \forall x \geq0$$
In this exercise, I'm supposed to use the mean value theorem. I always find myself having some trouble in these kinds of exercises. How can I prove this?
Equality holds for $x=1$, which suggests to use apply the mean value theorem to $f(x) - f(1)$ with $f(x) = \sqrt x$.
We get $$ \sqrt x - \sqrt 1 = (x-1) \frac{1}{2\sqrt c} $$ for some $c$ between $1$ and $x$. Now consider the cases $0 \le x < 1$ and $x > 1$ and show that the right-hand side is always $$ \le \frac{x-1}{2} \, . $$
(Note that we have essentially used is that the derivative of $f$ is decreasing. That is equivalent to $f$ being concave so that its graph lies below its tangent line at $x=1$.)
It is not necessary to use MVT. $$0\le(\sqrt x-1)^2=x-2\sqrt x+1 \iff 2\sqrt{x}\le x+1 \iff \sqrt{x}\le\frac{x+1}{2}$$ I use $x\ge0$ twice: first to be $\sqrt{x}$ well defined. Second, $(\sqrt{x})^2=|x|=x$.
For $x \ge 0$ we have:
\begin{align} \sqrt x \leq \frac{x+1}{2} &\iff x \le \frac{1}{4}(x+1)^2 \\&\iff 4x \le x^2+2x+1 \\&\iff 0 \le x^2-2x+1 \\&\iff 0 \le (x-1)^2. \end{align}
