# Proving inequality using mean value theorem

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, $$(\sqrt{x})^2=|x|=x$$.

• Yes, I know that. But in this exercise I'm asked to use the mean value theorem. – Eduardo Magalhães Nov 27 '20 at 10:22

For $$x \ge 0$$ we have:

$$\sqrt x \leq \frac{x+1}{2} \iff x \le \frac{1}{4}(x+1)^2 \iff 4x \le x^2+2x+1 \iff 0 \le x^2-2x+1 \iff 0 \le (x-1)^2.$$

• Yes, I know that. But in this exercise I'm asked to use the mean value theorem. – Eduardo Magalhães Nov 27 '20 at 10:22