If $xy=ab$ and $x<a\leq b<y$ then $x+y>a+b$.

Question

Let $a,b,x,y$ be positive integers $>0$. Suppose $$ \begin{align} xy&=ab,\\ x<a&\leq b<y \end{align}. $$ Then how to show that $x+y>a+b$?

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I saw this statement in a comment in the answer of @Haran to this question.

The cited question does not require the above statement to be true. It is sufficient if the statement with $\gcd(a,b)=1=x$ is true. In this case the statement is easy to see. But I have no idea how the general case can be deduced.

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Any ideas or hints are sincerely welcomed. Thanks in advance.

Answer 1

Since $x$ and $y$ are on the outside of $a$ and $b$, it is easy to see that $y-x>b-a\ge0$. Square both sides: $$y^2-2xy+x^2>b^2-2ab+a^2.$$ Now add $4xy=4ab$ to both sides, and take the square root of both sides.

Answer 2

This seems like an application of the AMGM inequality. That is, given two positive numbers whose product $P$ is known, minimize their sum. The minimum occurs at $\sqrt{P}$ and the sum increases from there. So it follows immediately from $x<a$, $y>b$.

Answer 3

Let the common product be $\tau$, so we have $y=\frac {\tau}x$.

For $x>0$, consider the function $$f(x)=x+\frac {\tau}x$$

and its derivative $$f'(x)=1-\frac {\tau}{x^2}$$

We see that the global minimum is attained when $x=\sqrt {\tau}$ and your desired claim follows immediately.

Answer 4

Just to give a different approach, note that $xy=ab$ implies $(xy)^N=(ab)^N$ for all powers $N$. From this, repeated squaring tells us

$$\begin{align} x+y\le a+b&\implies x^2+y^2\le a^2+b^2\\ &\implies x^4+y^4\le a^4+b^4\\ &\quad\vdots\\ &\implies x^{2^n}+y^{2^n}\le a^{2^n}+b^{2^n}\\ &\implies(x/y)^{2^n}+1\le(a/y)^{2^n}+(b/y)^{2^n}\\ &\implies\lim_{n\to\infty}((x/y)^{2^n}+1)\le\lim_{n\to\infty}((a/y)^{2^n}+(b/y)^{2^n})\\ &\implies0+1\le0+0\\ &\implies1\le0 \end{align}$$

which is, of course, a contradiction. So we must have $x+y\gt a+b$.