# Thread: Show that sqrt(1+x) <1+(1/2)x if x>0 ??

1. ## Show that sqrt(1+x) <1+(1/2)x if x>0 ??

I have another test question i got wrong and didnt undertand soo im hoping someone can help me out.

Show that sqrt(1+x) <1+(1/2)x if x>0 or ( formally written in attachment)

I used the formula f'(c)= [f(b)-f(a)]/(b-a) and then just ended up with a mess at the end which is probably why i got it wrong. How would i go abouts starting this problem?

2. Just square both sides

$\displaystyle 4+4x<4+x^2+4x$

$\displaystyle x^2>0$

3. Originally Posted by fobos3
Just square both sides

$\displaystyle 4+4x<4+x^2+4x$

$\displaystyle x^2>0$

Im a tad confused where the 4's came from.

4. Multiply both side by 2. $\displaystyle 2\sqrt{1+x}<x+2$

5. Originally Posted by kensington
I have another test question i got wrong and didnt undertand soo im hoping someone can help me out.

Show that sqrt(1+x) <1+(1/2)x if x>0 or ( formally written in attachment)

I used the formula f'(c)= [f(b)-f(a)]/(b-a) and then just ended up with a mess at the end which is probably why i got it wrong. How would i go abouts starting this problem?
If x is positive, prove

$\displaystyle \sqrt{1+x}<1+\frac{1}{2}x$

$\displaystyle 1+\frac{1}{2}x=\frac{2+x}{2}$

which is the positive square root of $\displaystyle \frac{4+4x+x^2}{4}$

Taking it for granted that we are referring to the positive square root of 1+x, then

$\displaystyle \sqrt{1+x}<\sqrt{\frac{x^2+4x+4}{4}}\ ?$

$\displaystyle \sqrt{\frac{4x+4}{4}}<\sqrt{\frac{x^2+4x+4}{4}}\ ?$

Yes

6. You can also use calculus to prove the inequality.

Consider the function $\displaystyle g(x)=\sqrt{x}-(1+\frac{1}{2}x)$. It suffices to show that when $\displaystyle x>0$, we have $\displaystyle g(x)>0$. First check $\displaystyle g(0)=0$ and find $\displaystyle g'(x)$ and show that it is always positive. Then you can conclude.

7. You'd need a variation on that.
If you prove a derivative is always positive,
then you are proving the function is always increasing.

$\displaystyle g(x)=(1+x)^{0.5}-1-0.5x$

Locate the maxmum value of g(x)

If $\displaystyle g'(x)=0\ for\ x=k$

then if g(k)<0 the inequality is proven.

Alternatively

$\displaystyle h(x)=1+0.5x-(1+x)^{0.5}$

Find the value of x=k giving minimum h(x),
then show h(k)>0

$\displaystyle 0.5-0.5(1+x)^{-0.5}=0$

$\displaystyle \frac{1}{\sqrt{1+x}}=1$

x=0.

2nd derivative test at x=0.....$\displaystyle 0.25(1+x)^{-1.5}$

this is 0.25 at x=0, indicating a minimum on the curve.

h(0)=0

This means that if x is positive, h(x)>0

8. Originally Posted by Archie Meade
You'd need a variation on that.
If you prove a derivative is always positive,
then you are proving the function is always increasing.
We also require $\displaystyle g(0)=0$, which I mentioned in my post.

For a more formal conclusion: we have $\displaystyle g(0)=0$ and $\displaystyle g(x)>0$ for all $\displaystyle x>0$. Then, by the Mean Value Theorem, there is $\displaystyle c\in (0, x)$ such that $\displaystyle \frac{g(x)-g(0)}{x-0}=g'(c)$. But this simplifies to $\displaystyle \frac{g(x)}{x}=g'(c)>0$. Multiplying $\displaystyle x$ to both sides gives the desired result.

9. Originally Posted by roninpro
We also require $\displaystyle g(0)=0$, which I mentioned in my post.

For a more formal conclusion: we have $\displaystyle g(0)=0$ and $\displaystyle g(x)>0$ for all $\displaystyle x>0$. Then, by the Mean Value Theorem, there is $\displaystyle c\in (0, x)$ such that $\displaystyle \frac{g(x)-g(0)}{x-0}=g'(c)$. But this simplifies to $\displaystyle \frac{g(x)}{x}=g'(c)>0$. Multiplying $\displaystyle x$ to both sides gives the desired result.
Thats the formula i mentioned before but this whole g(x) thing is confusing me

10. What part of $\displaystyle g(x)$ is causing you trouble?

11. Originally Posted by roninpro
What part of $\displaystyle g(x)$ is causing you trouble?
So im looking at my original question with the square roots and inequality and i dont see how i can take those values and sub then in the mean value Theorem.
I basically wote it as [1+(1/2)x - sqrt(1+x)]/ [(1/2)-(1+x)^(1/2)]... then eaither i got that wrong or i messed something up in simplifing because i couldnt prove thats its >0

12. Do you want to analyse that with the mean value theorem?

An algebraic or graphical solution is quite basic..

$\displaystyle f(x)=\sqrt{1+x}$

$\displaystyle g(x)=\frac{2+x}{2}$

$\displaystyle \sqrt{1+x}=\sqrt{\frac{4+4x}{4}}$

$\displaystyle \sqrt{\frac{4x+4}{4}}=\sqrt{\frac{x^2+4x+4}{4}}$

only when x=0.

Hence if x is positive, and since both functions are increasing functions

$\displaystyle f'(x)=\frac{1}{2\sqrt{1+x}}$

$\displaystyle g'(x)=\frac{1}{2}$

For x>0, g'(x)>f'(x) which means that after x=0, g(x) is increasing more than f(x)
for all x>0.

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