How to prove $\sqrt{1+x} \le 1+\dfrac{x}{2}$, $x > -1$ using MVT

Question

How to prove this question?(picture) I don't know the way and sorry for weird typing (i'm korean...)

Answer 1

Consider $f(t) = \sqrt{t+1} - \dfrac{t}{2} - 1$ on $(0,x)$, and take derivative $f'(t) = \dfrac{1}{2\sqrt{1+t}}- \dfrac{1}{2}$ . Now use the MVT: $f(x) - f(0)=\sqrt{x+1}-\dfrac{x}{2} - 1= (x-0)f'(t)< 0$ since $x > 0, f'(t) < 0$. If $x < 0$, then consider $f(t)$ on $(x,0)$ and we have $f(0) - f(x) = (0-x)f'(t)\implies f(x) = xf'(t)< 0$ since $x < 0, f'(t) > 0$. In both cases, $f(x) < 0$, and the result follows.. If $x = 0$ we have equality.

Answer 2

Let’s write down the MVT statement

$f(a)-f(b) = (a-b)f’(c)$ for some $c\in (a,b)$

for $a=x>b$ and $b=- \frac12$

assuming $f (x)=\sqrt{1+x}-1-\frac{x}{2}$

Answer 3

Consider $(1+(x/2))^2 = 1+ (x^2/4) + x \ge 1+x $. Thus $\sqrt{1+x} \le 1+(x/2)$.

Answer 4

Just square and you are done - using the mean value theorem is overkill!