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

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

1. 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 $(a,b)$?

2. 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
• Why do you keep on flipping the inequality?? – Nikunj Feb 28 '16 at 17:47
• 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

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.

• What is the meaning of "A.M-G.M" ? – Tharindu Sathischandra Mar 4 '16 at 10:45
• Arithmetic mean $\geq$ geometric mean – Nikunj Mar 4 '16 at 10:59

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.

• How we can show that $a^2(b-2013)+b^2(a-2013)=0$ has no solutions? – Tharindu Sathischandra Mar 4 '16 at 11:01
• There is a solution. Hint: Think about what happens if $b-2013=0$ and $a-2013=0$. – wythagoras Mar 4 '16 at 18:47
1. 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.