Is there a finite number of $(a, b)$ pairs that satisfy, ${a+b} \leq \frac {{a^2}b+{b^2}a}{a^2 +b^2}$

Question

${a+b} \leq \frac {{a^2}b+{b^2}a}{a^2 +b^2}$

Suppose that $a,b$ are two positive integers that satisfy the above equation. How can we show that there is a finite/infinite number of pairs for $(a,b)$?
Can there be two positive integers $(a,b)$ such that $\frac {{a^2}b+{b^2}a}{a^2 +b^2} =2013$ ; and how can we show that?

@Lovsovs I think you have reversed the inequality while editing it. — SchrodingersCat, Feb 28 '16 at 17:33
@SchrodingersCat I noticed that, but it was already the way I ended up doing it in the title, so I went with one of them. Do you think it was the other way around? (EDIT: Nope, I actually reversed both of them, sorry!) — Bobson Dugnutt, Feb 28 '16 at 17:34
Actuallty I reversed it (again), to make it match the original question. — wythagoras, Feb 28 '16 at 17:35
The reversed inequality that @Lovsovs edited is the correct one I needed. And I edited it again as Lovsovs did at the first time . — Tharindu Sathischandra, Feb 28 '16 at 17:47

Nikunj · Accepted Answer · 2016-02-28T17:55:07.823

1

Note that your expression can be written as $$ \frac {{a^2}b+{b^2}a}{a^2 +b^2}=\frac{ab(b+a)}{b^2+a^2}$$ Now $$ \frac{ab(b+a)}{(b^2+a^2)} \geq a+b$$ As $a,b$ are positive integers, $a+b \neq 0$ $$\implies a^2+b^2 \leq ab -(1)$$ Applying A.M-G.M on $a^2$ and $b^2$, we get $$a^2+b^2 \geq 2ab -(2)$$

This implies (1) is never true.

edited Feb 28 '16 at 17:55

answered Feb 28 '16 at 17:38

Nikunj

6,160

What is the meaning of "A.M-G.M" ? – Tharindu Sathischandra Mar 04 '16 at 10:45
Arithmetic mean $\geq$ geometric mean – Nikunj Mar 04 '16 at 10:59

wythagoras · Answer 2 · 2016-02-28T17:45:00.970

0

Hint: If we multiply both sides with $a^2+b^2$, which is positive, we get:

$$a^3+b^3+ab^2+a^2b=(a+b)(a^2+b^2)\geq ab^2+a^2b$$

For the second one, again multiply both sides with $a^2+b^2$ to get

$$a^2b+b^2a=2013a^2+2013b^2$$

We can rewrite this as $a^2(b-2013)+b^2(a-2013)=0$.

Now it shouldn't be to hard to find the solution.

edited Feb 28 '16 at 17:45

answered Feb 28 '16 at 17:37

wythagoras

25,026

How we can show that $a^2(b-2013)+b^2(a-2013)=0$ has no solutions? – Tharindu Sathischandra Mar 04 '16 at 11:01
There is a solution. Hint: Think about what happens if $b-2013=0$ and $a-2013=0$. – wythagoras Mar 04 '16 at 18:47

score 0 · Answer 3 · answered Feb 28 '16 at 18:10

0

substitute $x=ab,$ $y=a+b $ then you obtain equation $$2013 y^2 -xy -4026 x=0 $$ and on this site https://www.alpertron.com.ar/QUAD.HTM you can find a solutions of the above equation.

answered Feb 28 '16 at 18:10

Is there a finite number of $(a, b)$ pairs that satisfy, ${a+b} \leq \frac {{a^2}b+{b^2}a}{a^2 +b^2}$

3 Answers3

Linked