Most elegant proof for: $xy>ab$ does not imply $x+y>a+b$?

Question

Let $x,y,a,b$ be some constant real numbers.

Q1: Prove that $xy > ab$ does not imply that $x+y > a+b$ in the most elegant way?

Q2: What criteria do mathematicians use to define elegance?

My attempt:

Suppose it does imply that. Let:

• $x=2$
• $y=2$
• $a=4$
• $b=0.5$

Then while the 1st inequality is true: $$\begin{split} xy&>ab\\ 2\times 2&>4\times 0.5\\ 3&>2\\ \end{split}$$ the second is false: $$\begin{split} x+y&>a+b\\ 2+ 2&>4+ 0.5\\ 4&>4.5\\ \end{split}$$

And that's hopefully a proof by contradiction.. or is it? I suspect that it's nasty and that I am doing some jumps somethere.

I would really appreciate guidance to reach a proof for this simple problem that is most rigor and most elegant.

Answer 1

A single counterexample suffices. Put $$x=y=3\ ,\qquad a=8,\quad b=1\ .$$ Then $xy>ab$, but $x+y<a+b$.

Answer 2

Let $A = (a, b)$ and $X = (x, y)$ be points in the first (positive $X, Y$) quadrant of the $X-Y$ plane with $x, y \ne 0$. Let $R_A$ and $R_X$ be the rectangles formed by parallels to the axes to these points (e.g. $R_A $ has vertices $(0, 0), (0, b), (a, b), (a, 0))$.

For any finite values $w, z$ we can choose $A $ with length greater than $w$ and area $R_A < z$ by moving the point $A$ far along one axis and close to the other.

Therefore we can choose a point $A$ with $a > x + y$ and $ab < xy$, and since $b $ is also positive then $a + b > x + y$ and $ ab < xy$.

(i.e. $xy > ab$ and $x + y \not \gt a + b$ )

P.S. - I think that is elegant.