The following is problem 16 in chapter 2 from Spivak's book:
Prove that if $m$ and $n$ are natural numbers and $\frac{m^2}{n^2} \lt 2$, then $\frac{(m+2n)^2}{(m+n)^2} \gt 2$; show, moreover, that $$\frac{(m+2n)^2}{(m+n)^2} -2 \lt 2- \frac{m^2}{n^2}.$$
I understand the first demonstration, but I don't understand the second one.
I saw the solution book, but only the last expression is developed there, and I do not understand how that shows what is asked.
Thanks.
On
I know it's been a while since this question was asked, but for future readers here's a detailed this. Basically, we need to show that:
$$\frac{m^2}{n^2}<2 \implies 0<\frac{(m+2n)^2}{(m+n)^2}-2< 2 - \frac{m^2}{n^2}$$
for natural numbers $m$ and $n$
Since the original question asks only for help with the second part of the question I will prove the inequality $\frac{(m+2n)^2}{(m+n)^2}-2< 2 - \frac{m^2}{n^2}$ first and then add the proof of the inequality $0<\frac{(m+2n)^2}{(m+n)^2}-2$ (which is of course the same as the inequality $\frac{(m+2n)^2}{(m+n)^2}>2$) for interested readers at the end.
To start off with, consider the expression $\frac{(m+2n)^2}{(m+n)^2}-2$. Simplifying this, we have
$$\frac{(m+2n)^2}{(m+n)^2}-2 = \frac{([m+n] + n)^2}{(m+n)^2}-2$$
$$ = \frac{(m+n)^2 + n^2 + 2n(m+n)}{(m+n)^2}-2$$
$$ = 1 + \frac{n^2 + 2n(m+n)}{(m+n)^2}-2 = \frac{n^2}{(m+n)^2} + \frac{2n}{(m+n)} -1 $$
$$= \frac{2n^2}{(m+n)^2} - \frac{n^2}{(m+n)^2} + \frac{2n}{(m+n)} -1$$
$$= \frac{2n^2}{(m+n)^2} - \left[ \frac{n^2}{(m+n)^2} + 1 - \frac{2n}{(m+n)} \right]$$
$$= \frac{2n^2}{(m+n)^2} - \left[ 1 - \frac{n}{(m+n)} \right]^2 = \frac{2n^2}{(m+n)^2} - \left[ \frac{m}{(m+n)} \right]^2 $$
$$= \frac{(2n^2 - m^2)}{(m+n)^2} = \frac{n^2}{(m+n)^2}\left[ 2 - \frac{m^2}{n^2} \right]$$
Therefore, we have
$$\frac{(m+2n)^2}{(m+n)^2}-2 = \frac{n^2}{(m+n)^2}\left[ 2 - \frac{m^2}{n^2} \right]$$
But $(n)^2$ is obviously lesser than $(m+n)^2$ which implies that $\frac{n^2}{(m+n)^2}<1$ from which it follows that
$$\frac{(m+2n)^2}{(m+n)^2}-2 = \frac{n^2}{(m+n)^2}\left[ 2 - \frac{m^2}{n^2} \right] < 2 - \frac{m^2}{n^2}$$
This completes the required proof.
Note:
Since the inequality $\frac{m^2}{n^2}<2$ has not been mentioned anywhere in the above proof, one might be tempted to ask if the condition is even necessary for the final result. Taking examples where $\frac{m^2}{n^2}>2$, you'll begin to see that the answer is a resounding yes. But why? I suggest you ponder this question yourself, but in any case I'll include a justification for this here.
Justification:
In the final step we write $$\frac{n^2}{(m+n)^2}\left[ 2 - \frac{m^2}{n^2} \right] < 2 - \frac{m^2}{n^2}$$ Notice carefully that this inequality holds if and only if $\left[ 2 - \frac{m^2}{n^2} \right]$ is positive. If not, the inequality sign needs to be reversed. Thus the initial inequality that $\frac{m^2}{n^2}<2$ is indeed crucial for the final result to be true.
Additional Note: If you have found this problem in Spivak's Calculus, you'll see that the very next question asks for the proof of the same inequality but with all the inequality signs reversed. The above observation makes that proof follow rather easily from the above proof.
Below is a solution to the first part of the problem.
In this part we are to prove that $\frac{m^2}{n^2}<2 \implies \frac{(m+2n)^2}{(m+n)^2}>2$
We start with the expression $\frac{(m+2n)^2}{(m+n)^2}$ and proceed as follows:
$$\frac{(m+2n)^2}{(m+n)^2} = 1 + \frac{n^2 + 2n(m+n)}{(m+n)^2} = 1 + \frac{3n^2 + 2mn}{(m+n)^2} $$
But we've already assumed that $\frac{m^2}{n^2}<2$ from which it follows that
$$m^2<2n^2\implies m^2 + n^2 < 3n^2$$
Using this inequality in the above equation we get
$$\frac{(m+2n)^2}{(m+n)^2} = 1 + \frac{3n^2 + 2mn}{(m+n)^2} > 1 + \frac{m^2 + n^2 + 2mn}{(m+n)^2}= 2 $$
Thus we prove successfully that $\frac{(m+2n)^2}{(m+n)^2} > 2$
Solution:
$$\text{Simplifying }\frac{(m+2n)^2}{(m+n)^2} > 2$$
$$\text{we get } (m+2n)^2 > 2 (m+n)^2$$
$$\text{i.e. } m^2 + 4 n^2 + 4mn > 2 m^2 + 2 n^2 + 4mn$$
$$\text{i.e. } 2n^2 > m^2 \text { , which is true as per the given condition}$$