If $0 
Sophia Terry
Sophia Terry 
I first came across this problem on 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.)
$\endgroup$ 22 Answers
$\begingroup$$$(0<x<1\land 0<y<1)\implies$$ $$\implies (0<1-x\land 0<1-y)\implies$$ $$\implies0<(1-x)(1-y)\implies$$ $$\implies 0<1-(x+y)+xy \implies$$ $$\implies x+y<1+xy.$$
$\endgroup$ 2 $\begingroup$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.
$\endgroup$ 7
