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

#### 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?

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#### fobos3

Just square both sides

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

$$\displaystyle x^2>0$$

#### kensington

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.

#### fobos3

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

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)
View attachment 18197

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

kensington

#### roninpro

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.

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

#### roninpro

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.

#### kensington

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

#### roninpro

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