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So as the title states I have to show $\sqrt{1+x} < 1+\frac{x}{2}$ for $x>0$.

This is a example from the book which has to be explained for me, i'm having a hardtime understanding the proof. I do however understand the concept of MVT.

So the rest of the solution looks like the following:

If $x>0$, apply the Mean-Value Theorem to $f(x)= \sqrt{1+x}$ on the interval $[0,x]$. There exist $c\in [0,x]$ such that $$\frac{\sqrt{1+x}-1}{x}=\frac{f(x)-f(0)}{x-0}=f'(c)=\frac{1}{2\sqrt{1+c}}<\frac{1}{2} $$ The last inequality hold because $c>0$. Mulitiplying by the positive number $x$ and transposing the $-1$ gives $\sqrt{1+x} <1+\frac{x}{2}$,

So I am not sure why he(the author) choose $\frac{1}{2}$, it seems a little arbitrary to me. My guess is that you're allowed to pick a number for the derivative of $c$ which suits the cause/solution best, as long as it's $0<c<x$. I'm not sure though.

All help would be greatly appriciated!

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    $\begingroup$ MVT is definitely the hard way to prove the inequality. Heaps simpler to just square both sides! $\endgroup$
    – Gerry Myerson
    Nov 30, 2017 at 1:08

3 Answers 3

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$\frac{1}{2}$ is not a random number, it comes from differentiating the square root: $\sqrt{x}\,'=\frac{1}{2\sqrt{x}}$ or $\sqrt{x+1}\,'=\frac{1}{2\sqrt{x+1}}$, the same thing.

Next he noticed that $\frac{1}{\sqrt{1+c}}<\frac{1}{\sqrt{1+0}}=1$ for all $c>0$, which is quite obvious after simple transformation.

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  • $\begingroup$ I see. But what's the next step then. i.e the proof? Is the point then to make $\frac{1}{2sqrt{1+c}}<\frac{1}{2}$ look like $\sqrt{1+x} <1+\frac{x}{2}$? $\endgroup$
    – Aliasa Zarowny Pseudonymia
    Nov 30, 2017 at 1:37
  • $\begingroup$ He obtained the inequality $\frac{\sqrt{1+x}-1}{x}<\frac{1}{2}$. This one is equivalent to the inequality you were given to prove. $\endgroup$
    – tong_nor
    Dec 1, 2017 at 15:17
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Just square both sides.

$(1+x/2)^2 = 1+x+x^2/4 \gt 1+x$ unless $x = 0$.

Also works for $(1+x/n)^n \gt 1+x$ so $1+x/n \gt \sqrt[n]{1+x}$ (in this case, use Bernoulli's inequality).

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  • $\begingroup$ Yeah I'd like to, but im trying to learn how to prove inequalities using MVT. :/ $\endgroup$
    – Aliasa Zarowny Pseudonymia
    Nov 30, 2017 at 2:30
  • $\begingroup$ If $ c < 1/2$ then $(1+cx)^2 = 1+2cx+c^2x^2=1+x-(1-2c)x+c^2x^2 < 1+x$ for small enough $x$. Therefore $1/2$ is the smallest value that works. $\endgroup$
    – marty cohen
    Nov 30, 2017 at 2:48
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$\frac12$ is not arbitrary at all.

The derivative of a square root is $\sqrt{u}'=\frac{u'}{2\sqrt{u}}$ put $u=x+1$ you get $\frac{1}{2\sqrt{1+x}}$ replace the $x$ with the value from the mean value theorem you get $\frac{1}{2\sqrt{1+c}}$

Look at the case where he choose more than $\frac12$, in this case you lose some of the information, because if it is less then $\frac12$ then it is less from something bigger.

In the case he choose something smaller, then I can take $\lim\limits_{c\to0}\frac1{2\sqrt{1+c}}\longrightarrow\frac12$, which means I can get infinitely close to $\frac12$ hence I can find a value for $c$ that will make the statement false for every number less then $\frac12$

Thus we left with the number $\frac12$

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