How to prove $xy>x+y$ if $x,y \in (2,\infty)\subseteq \Bbb R$

Question

I am new to proofs. This is my first proof-based mathematics class, and it is a hard transition from my high school classes. I am a second semester freshman.

My original question was:

Prove that there is a real number with the property that for any two larger numbers, there is another real number that is larger than the sum of the two numbers and less than their product.

I have attempted this by a constructive proof (which is possible according to my distinguished professor) and am very unwilling to attempt a nonconstructive proof until I have proven it by a constructive proof. I said:

"Let $x,y,z,$ and $t \in \Bbb R$, with $y>x$ and $z>x$. Because $x,y,z,$ and $t$ are real numbers, if we choose a pair of $y$ and $z$ such that $yz>(y+z)$ , the following always holds $\exists t(yz>t>(y+z))$. Let $x=2$ and let $\epsilon$ be some arbitrary infinitesimal positive quantity. The smallest product of $y$ and $z$ would then be the product of $y=2+\epsilon$ and $z=2+\epsilon$ and the product would be $yz=4+2\epsilon + \epsilon^2$. Similarly, the smallest sum of $y$ and $z$ would be the sum if $y=2+\epsilon$ and $z=2+\epsilon$ which is $(y+z)=4+2\epsilon$. We see that $yz>(y+z)$ which is equivalent to $4+2\epsilon + \epsilon^2 > 4+2\epsilon$..."

And this is where I get stuck. I don't think I am near the end of the proof, I feel like I need a lemma to complete this. I want the lemma to be the proof of:

Prove that for any pairs of real numbers $x$ and $y$, where $x>2$ and $y>2$, the following always holds: $xy>(x+y)$

I thought of proving this lemma by approaching using the AM-GM inequality which I learned in competitive mathematics during high school, since one side is a product and one side is a summation, but I have not started it. I feel like I am being a little redundant in the proof, going in circles, or writing without much thinking.

Disregarding the original question (although I appreciate hints), how would I approach the proof of the lemma?

Answer 1

The inequality is equivalent to $$(x-1)(y-1)>1 $$ which clearly holds for $x,y>2$. Then, if you take $t=\frac{xy+x+y}{2}$, you will have $x+y<t<xy$.

Answer 2

If $x>z$ and $y > z$ then $a=x-z> 0$ and $b=y-z>0$.

And $x+y = (z+a) + (z+b) = 2z +(a+b)$. And $xy = (z+a)(z+b) = z^2 + (a+b)z + ab$.

We want to find a $z$ so that $2z + (a+b) < z^2 +(a+b)z + ab$ for all possible positive $ab$.

This will happen if and only if $0 < (z^2-2z) + (a+b)(z-1) + ab=z(z-2) + (a+b)(z-1) + ab$ for all positive $a,b$

We know that $ab > 0$ and $(a+b)>0$ for all positive $a,b$ so if we can have $z(z-2)\ge 0$ and $z-1\ge 0$ that will be good enough.

If $z = 2$ and $x>2$ and $y > 2$ and $a= x-2>0;b=y-2>0$ then $xy=(2+a)(2+b) = 4+2(a+b)+ab> 4+a+b = x+y$.

So $z=2$ is just such a number.

And if $x+y < xy$ then $w=\frac {(x+y) + xy}2$ will be that $x+y < w < xy$ and $w$ is such a number as we need.

Note: We haven't proven that $z=2$ is the least such possible value but we weren't asked to prove that.

But we can prove that.

If $z < 2$ we just have to prove there exist any $x,y> z$ where this fails.

If $x = y = 1$ then $x+y = 2 >1=xy$ so this will fail if $z< 1$. ANd if $1\le z< 2$ then if we let $x+y = \frac {z+2}2$. Then $x+y = z+ 2$ while $xy= \frac{z^2 + 4z + 4}{4} = \frac {z^2}4 + z + 1= z + (\frac {z^2}4 +1)$. But $1\le x < 2$ so $z+(\frac {z^2}4 +1) < z + (1+1)=z+2 =x+y$. So that is a counter example.

So $z$ is the smallest such number. (Similar for any $z > 2$ this will hold. After all, if $x,y > z > 2$ then $x,y > 2$ and we know that in that case $x+y < xy$.)

Answer 3

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

Since ${1\over x-1}$ is decreasing for $x\geq 2$ it has maximum value at $x=2$, so for $x>2$ it is less than $1$ and thus $y>2$.