Using the mean value theorem prove the below inequality.

$$\frac{1}{2\sqrt{x}} (x-1)<\sqrt{x}-1<\frac{1}{2}(x-1)$$ for $x > 1$.

I don't understand how these inequalities are related. Am I supposed to work out the first one and then the second and so on? I also would be really grateful if anyone had the time to give some insight in what this problem asks to me.

I really wish someone could give a very simple solution.


Apply the mean value theorem to the function $f(t) = \sqrt{t}$ on the interval $[1,x]$ to deduce

$$\sqrt{x} - 1 = \frac{1}{2\sqrt{c}}(x - 1)$$

for some $c \in (1,x)$. Use the fact that

$$\frac{1}{2\sqrt{x}} < \frac{1}{2\sqrt{c}} < \frac{1}{2}$$

to conclude.

  • $\begingroup$ how did you know $$f(t)=\sqrt{t}$$? can you show me a different approach too? if possible. $\endgroup$ – Sherlock Homies May 30 '15 at 21:27
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    $\begingroup$ @SherlockHomies note $\sqrt{x} - 1 = \sqrt{x} - \sqrt{1} = f(x) - f(1)$, where $f(t) = \sqrt{t}$. If you want to prove the inequalities without the use of MVT, then rationalize the numerator to get $$\sqrt{x} - 1 = \frac{x-1}{\sqrt{x} + 1}$$ and use the fact that for $x > 1$, $$\frac{1}{2\sqrt{x}} < \frac{1}{\sqrt{x} + 1} < \frac{1}{2}.$$ $\endgroup$ – kobe May 30 '15 at 21:30
  • $\begingroup$ wait a sec how would you piece up all together to get to end of the problem. Like in the question above . I mean the last step.Using the mean value theorem again in the end? $\endgroup$ – Sherlock Homies May 30 '15 at 21:49
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    $\begingroup$ @SherlockHomies multiply the inequalities $\frac{1}{2\sqrt{x}} < \frac{1}{2\sqrt{c}} < \frac{1}{2}$ by $x - 1$ and use the fact $\sqrt{x} - 1 = \frac{1}{2\sqrt{c}}(x - 1)$ to get the desired inequalities. $\endgroup$ – kobe May 30 '15 at 21:52

By the MVT, there is $c \in ]1,x[$ such that $$\sqrt{x} - 1 = \frac{1}{2\sqrt{c}}(x-1), \qquad \qquad \left[f(b)-f(a) = f'(c)(b-a)\right]$$ so you use that: $$c < x \implies \sqrt{c} < \sqrt{x} \implies 2 \sqrt{c} < 2\sqrt{x} \implies \frac{1}{2\sqrt{x}}<\frac{1}{2\sqrt{c}}$$ to get one side, and use that: $$c > 1 \implies \sqrt{c} > 1 \implies 2\sqrt{c} > 2 \implies \frac{1}{2\sqrt{c}}< \frac{1}{2}$$ to get the other.


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