prove inequality through mean value theorem

Solution 1:

Here's the essential idea:

Say we want to show that $f(x) \leq g(x)$ over some interval $I$ (with $f$ and $g$ being both continuous and differentiable). Then define $h(x) = g(x)-f(x)$; the aim is to show that $h$ is always nonnegative.

From the Mean Value Theorem, we have for some $a, y, c \in I$ (with $a \leq c \leq y$) that $$\min_{a \leq x \leq y} h'(x) \leq h'(c) = \dfrac{h(y)-h(a)}{y-a} \leq \max_{a \leq x \leq b} h'(x)$$

Then if we can show that $\displaystyle\min_{a \leq x \leq y} h'(x) \geq 0$, then clearly we must have that $h(y) \geq h(a)$ for all $y \geq a$, or that $g(y) \geq f(y)+g(a)-f(a)$ for all $y \geq a$.

After this, generally the premise of the question makes it easy to verify that $g(a) \geq f(a)$, and you should be set (in this case, setting $a = -1$ sets the premise).

Solution 2:

Take $f(x) = \sqrt{1+x}-1-x/2$. What you want to do is to show that, using the Mean Value Theorem, that whenever $x > -1 $, $f'(x)$ is positive. What does this tell us? That $f(x)$ is increasing when $x > -1$. Now, if we have $f(x) \geq 0$ on our interval for some $x_0$, we know that $f(x) \geq 0$ for every $x \geq x_0$. Thus, we get that $$\sqrt{1+x}-1-x/2 \geq 0$$ or $$\sqrt{1+x} \geq 1+x/2$$ Do you see why? Since $f(x)$ is increasing, it never goes back across zero.

This is how you apply the MVT to prove inequalities. But, as Steven said, this inequality in particular isn't true.

Solution 3:

You can simply choose $f$ as the left-hand side of the inequality.

The mean-value theorem, applied to $f(x) = \sqrt{1+x}$, states that for some $c$ between $0$ and $x$, $$ f(x) = f(0) + x f'(c) = 1 + \frac{x}{2\sqrt{1+c}} \, $$ and the last expression is always $$ \le 1 + \frac x2 \, . $$