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A) Use the Mean value theorem to prove that \begin{equation} \sqrt{1+x} < 1 + \frac{1}{2}x \text{ if } x>0 \end{equation}

B) Use result in A) to prove that \begin{equation} \sqrt{1+x}>1+\frac{1}{2}x-\frac{1}{8}x^2 \text{ if } x>0 \end{equation}

Can someone give an answer for part B) ?

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    Do you absolutely need MVT for a? It can be shown in a simpler way – 2017-01-03
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    @Alex But often such examples are used when the point is not to solve them in the easiest way, but as exercises on using newly explained results. – 2017-01-03
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    @Alex can you show your method for A) without using Mean Value Theorem? – 2017-01-03
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    @yh05 If you are interested in an easy way to show (A), take the well known inequality $\frac{a^2 + b^2}{2}\geq ab$ which holds true for real $a,b$ and set $a=\sqrt{x+1}$ and $b=1$. The equality does not hold, because at this case we would have $a=b$ in the initial inequality and thus we would also have $\sqrt{1+x}=1\Rightarrow x=0$, which is not acceptable. – 2017-01-03
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    @SachpazisStelios thanks. i didn't know of this inequality. – 2017-01-03
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    @yh05 The $\frac{a^2 + b^2}{2}\geq ab$ can be derived from $(a - b)^2\geq 0$ by adding $2ab$ to both sides and dividing by $2$ – 2017-01-03
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    @Blauelf Yes, i noticed it as well. – 2017-01-03

2 Answers 2

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From (A), for $x>0$ we have:

$0<\sqrt{x+1}-1<\frac{1}{2}x\Rightarrow {(\sqrt{x+1}-1)}^2<\frac{1}{4}x^2\Rightarrow {(\sqrt{x+1})}^2+1^2-2\sqrt{x+1}<\frac{1}{4}x^2 \Rightarrow \sqrt{x+1}>1+\frac{x}{2}-\frac{x^2}{8}$

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    Thanks alot. I took quite a long time to attempt the question before you posted your solution :( How do you thought of this solution? – 2017-01-03
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    @yh05 I observed the $x^2$ term and so I thought that I should square both sides of an inequality, equivalent to the one I already had in (A). Therefore I squared both sides of the inequality, after taking 1 to its left hand side. The motivation to move 1, was that the square of the $\sqrt{x+1}-1$ would give me a $-2\sqrt{x+1}$ and I needed the negative sign to reverse the inequality. – 2017-01-03
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OK, I'll show how I'd solve a:

We have

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

Clearly $f(0) = 0$. Let's look at the derivative of $f$. If we show that the derivative is strictly positive for $x>0$, the function is strictly increasing, hence positive (in combination with $f(0)$) for $x>0$. We have

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

Since $x>0$, the term in the brackets is always positive, hence the derivative is strictly positive and the function is increasing. Hence,

$$ f(x)>0 \to 1+\frac{x}{2} > \sqrt{1+x} $$

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    Yes this method is definitely more straightforward than applying MVT. – 2017-01-03
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    yes thanks I fixed it. This is a useful approach in general – 2017-01-03
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    I'm pretty sure the easiest way to solve part (a) if you don't have to use MVT is just to square both sides. – 2017-01-03
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    That's the second easiest – 2017-01-03