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Find the value of $$\frac{a^3+b^3+c^3}{abc}\qquad\text{ if }\quad \frac ab + \frac bc + \frac ca = 1.$$

I tried using Cauchy's inequality but it was of no help. Please guide me.

$a, b, c$ are real.

  • 1
    are $a,b,c$ real, complex? They cannot be all positive...2012-06-29
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    a, b, c are real.2012-06-29
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    Are you sure this is the question, because I'm not sure there is enough information to compute the value. In particular, if $(u,v,w)=(a/b,b/c,c/a)$ then $uvw=1$ and $u+v+w=1$ but the expression you are seeking is not a symmetric function in $(u,v,w)$, and so cannot be determined just by knowing $u+v+w=1$ and $uvw=1$.2012-06-29
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    @ThomasAndrews: I believe he wants the maximum value.2012-06-29
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    @EricNaslund I was sort of guessing that, after the fact, especially since he tried using an inequality, but that is not what he asked, which was why I asked if that was really the problem he was trying to solve.2012-06-29
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    @ThomasAndrews: I wanted to add, even if it _was_ a symmetric function, its evaluation would still not be possible. At the moment, the polynomial is $u^{2}v+v^{2}w+w^{2}u$, which is not symmetric, but even if it was was $u^{2}v+v^{2}u+v^{2}w+w^{2}v+w^{2}u+u^{2}w,$ we still would not be able to evaluate it exactly.2012-06-29
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    Yeah, but it *might* be possible if it was symmetric, it might not. Without being symmetric, though, it is clearly not possible. :) @EricNaslund2012-06-29
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    Maybe he wants something like $\frac{a^3+b^3+c^3}{abc}=\frac{(a+b+c)(a^2+b^2+c^2-ab-ac-bc)}{ab^2+a^2c+bc^2}+3$.2012-06-29
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    Maybe working out some cases might help: Case 1: Suppose $b=k_1$ and $c=-k_2$ where $k_1,k_2>0$. Then we have $\frac{a}{k_1}-\frac{k_1}{k_2}-\frac{k_2}{a}=1\implies a=\dfrac{k_1+\frac{k_1^2}{k_2}\pm \sqrt{(k_1+\frac{k_1^2}{k_2})^2+4k_1k_2}}{2}.$ Since $a$ is real , we know that discriminant here is nonnegative and maybe guess some values for $k_1,k_2$.2012-06-29
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    I am **certain** that this is a duplicate.2012-06-30

3 Answers 3

1

There is not enough information to solve this problem. Clearing out denominators, your hypothesis is $$a^2 c + a b^2 + b c^2 = abc \quad (1)$$ and your desired conclusion is $$a^3+b^3+c^3=kabc \quad (2)$$ for some constant $k$.

Suppose, for the sake of contradiction, there were a $k$ such that $(1)$ implied $(2)$. Since the polynomial $a^3+b^3+c^3-abc$ is irreducible, this would mean that $a^2 c+a b^2+b c^2-abc$ would divide $a^3+b^3+c^3-kabc$. But the two polynomials are both cubics, so the only way for the first to divide the second is the first is a scalar multiple of the second, and it isn't.

It's also easy to generate points on $(1)$ and see that the ratio $(a^3+b^3+c^3)/(abc)$ isn't constant. Just choose random values for $a$ and $b$ and equation $(1)$ turns into a quadratic; solving that quadratic gives you some points to try. You'll see very quickly that nothing like this is true.

0

Let

$$\begin{eqnarray} u&=&a+b+c\\ v&=&a^3+b^3+c^3\\ s&=&a^2b + b^2 c + c^2 a\\ t &=& a \, b\, c \end{eqnarray}$$

These are related by $$u^3 = v + 3 s + t $$

Further, we are given

$$ \frac ab + \frac bc + \frac ca = \frac st =1$$

so $$u^3 = v + 4 t$$

And our target is

$$\frac{a^3+b^3+c^3}{abc}=\frac vt = \frac{u^3}{t}-4$$

This cannot be a single value, as we have a remaining degree of freedom.

-2

Take $a=1$ and $b=-\epsilon$. Then $c=O(1/\epsilon)$ ($\epsilon\rightarrow 0$). By taking $\epsilon$ suitably small, we can make the expression to be evaluated, of order $1/\epsilon^3$, big or much bigger.