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

  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?

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  • $\begingroup$ @Lovsovs I think you have reversed the inequality while editing it. $\endgroup$ – SchrodingersCat Feb 28 '16 at 17:33
  • $\begingroup$ @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!) $\endgroup$ – Bobson Dugnutt Feb 28 '16 at 17:34
  • $\begingroup$ Actuallty I reversed it (again), to make it match the original question. $\endgroup$ – wythagoras Feb 28 '16 at 17:35
  • $\begingroup$ Why do you keep on flipping the inequality?? $\endgroup$ – Nikunj Feb 28 '16 at 17:47
  • $\begingroup$ The reversed inequality that @Lovsovs edited is the correct one I needed. And I edited it again as Lovsovs did at the first time . $\endgroup$ – Tharindu Sathischandra Feb 28 '16 at 17:47
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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.

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

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