2
$\begingroup$

I first came across this problem on http://nrich.maths.org/ and have tried to solve it myself. Here is what I did (may seem laughable to those who know the answer!):

Let $a = x+y$ and $b=1+xy$

I figured that to prove $x+y<1+xy$ to be correct, I need to find an equation which will allow me to plot a graph, $d = b-a$, showing that $b - a$ approaches 0, but never touches it. This is where I am stuck and don't know how to proceed. Presumably, if my method has been correct so far, I need to simplify it in some way so that there is one variable only.

(I haven't really been taught anything like this at school but since I want to study maths later on, I figured it may be helpful to try some challenging questions like this. This seemed fairly easy at first, but can't get my head around it, so any helpful answer which I can learn from is greatly appreciated.)

  • 8
    $(1-x)(1-y) = (1+xy) - (x+y)$. Can you proceed from here? – 2017-01-18
  • 1
    Alternatively, a much less obvious way is to utilize the addition formula $$ \tanh(a+b) = \frac{\tanh a + \tanh b}{1+\tanh a \tanh b} $$ together with the fact that the range of $\tanh$ is $(-1, 1)$. – 2017-01-18

2 Answers 2

2

We start by rewriting the expression:

$$x + y < 1 + xy \iff x + y - xy < 1 \iff x(1 - y) + y < 1$$

Assume $y$ is fixed. How much would $x$ have to be in order to make that an equality? i.e. let us solve for $x$ the equation

$$x(1-y) + y = 1 \iff x(1-y) = 1-y \iff x = \frac{1-y}{1-y} = 1$$

where the division can be done because $y \not=1 \implies 1-y \not= 0$.

Therefore if $x = 1$, we would have $x(1-y) + y = 1$. Decreasing $x$ will decrease the value of $x(1-y) + y$ given that $1-y > 0$ and thus we have $x(1-y) + y < 1$ as we wanted to show.

  • 0
    Thank you for laying it out in a really understandable way, that's really fascinating. I didn't ever think about rearranging the inequalities. So, if I understand correctly, as long as $0 < x < 1$, then $x(1−y)+y$ will also be between 0 and 1. $y$ cannot be 1 because were are told that y cannot be more than or equal to 1, so $x$ will never be 0. And x will cannot be more than 1 because when $x = 1$, $x(1−y)+y$ will equal 1, when it should be <1. – 2017-01-18
  • 0
    @Sangchul Lee posted a hint above (1st comment). I was unable to follow it, but if I were to, would it have been the same proof as the one you typed? Thanks. – 2017-01-18
  • 0
    @FrankShang you got it right until the middle; After we rewrite the inequality, we want to show that $x(1-y) + y < 1$. We don't know that yet. But we reason like this: $x(1-y) + y$ would be $1$ if $x$ was equal to $1$. But we know that $x < 1$, so we have that $1(1-y) > x(1-y)$ and therefore $1 = 1(1-y) + y > x(1-y) + y$ which shows what we wanted to show – 2017-01-18
  • 0
    @FrankShang it would not have been the same, no, it is just another way of approaching the problem. And actually a very nice one – 2017-01-18
  • 1
    Ah, that suddenly became really clear! Thank you for taking the time to reaffirm that. I know this is unrelated, but are proofs like these expected from someone who is still a few years away from attending university, even if they are very enthusiastic about maths? – 2017-01-18
  • 0
    @FrankShang In mathematics, and specially for proofs, one often need a bit of intuition so as to find the path that one should follow. That comes with practice and expertise; the more proofs you write - of all sorts - the more easy it will become. Same goes with (numerical) problems. The more experience one gets, the easier it is to the find the better approach. – 2017-01-19
  • 1
    That's very encouraging thank you. I hope that in 5-6 years I could be answering some of the questions on this site! – 2017-01-19
3

$$(0

  • 0
    I have a feeling this is the solution @Sangchul Lee had in mind. Am I correct? That is a very concise but clear solution. Thank you! – 2017-01-19
  • 0
    @FrankShang it is indeed – 2017-01-19