prove inequality through mean value theorem I want to prove that if $x$ is a real number and $x>-1$,
$$\sqrt{1+x}  \leq 1+\dfrac{x}{2}$$
I'm not sure which one should I choose for $f(x)$. I also don't understand why we need the mean value theorem to prove this.
 A: 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).
A: 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.
A: 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
$$
 \le 1 + \frac x2 \, .
$$
A: Notice  by the AM-Gm inequality: $y \geq 0 $,
$$ \frac{y + 1}{2} \geq \sqrt{y} $$
Now, put $y = x+1 $. Then for $x \geq -1 $,
$$ \frac{ (x +1) + 1}{2} \geq \sqrt{x+1} \implies \frac{x}{2} + 1 \geq \sqrt{x+1}$$
No need to use mean value theorem.
