Maximum Product of Sums

The first step is to multiply out the expression. First lets' look at just the distribution of $\frac{a}{b-c}$ : $\frac{a}{b-c} (\frac {b-c}{a} + \frac {c-a}{b} + \frac {a-b}{c}) = 1 + \frac{a}{b-c} ( \frac{c^{2} - ac + ab - b^{2}}{bc}) = 1 + \frac{a(a-c-b)}{bc} = 1 + \frac {2a^2}{bc}$ . Similarly, the distributions for the second and third terms will be $1+\frac{2b^{2}}{ac}$ and $1+\frac{2c^{2}}{ab}$ respectively, because the expression is symmetrical. Therefore, the entire expression evaluates out to $3+(\frac{2a^{2}}{bc}+\frac{2b^{2}}{ac}+\frac{2c^{2}}{ab})$ which equals $3+2(\frac{a^{3}+b^{3}+c^{3}}{abc})$ .

To simplify this, we use the factorization $a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$ , hence $a^{3}+b^{3}+c^{3}=3abc$ . Now we can substitute this back into our expression, and we get $3+2(\frac{3abc}{abc})$ which equals $3+6=9$ . Therefore, the expression always equals $9$ so the maximum value is $9$

a+b+c = 0; therefore replace c with -(a+b) in the given equation. the given equation would turn up in terms of a and b. Precisely, ((a/(2b+a)) - (b/(2a+b)) - ((a-b)/(a+b))) * (((2b+a)/a) - ((2a+b)/b) - ((a-b)/(a+b))).

Now divide the numerator and denominator part by b. the equation would convert in terms of a/b. Replace a/b with x. Now you have an equation in one variable.

On simplifying then onwards you will get (-9x(x+1)/((x-1)(2x+1)(2+x))) * (-((2+x)(2x+1)(x-1))/(x(x+1)))

Eventually everything would get cancelled and 9 will remain. Thus for the given problem it will have a solution of 9.

We have (abc) \neg 0 and ((a-b)(b-c)(c-a)) \neg 0

Let P = \frac {a}{b-c} + \frac {b}{c-a} + \frac {c}{a-b}

P(b-c)(c-a)(a-b) = a(c-a)(a-b) + b(b-c)(a-b) + c(b-c)(c-a) = a(c-a)(a-b) - (a+c)(b-c)(a-b) + c(b-c)(c-a) = a(c-a)(a-b) - a(b-c)(a-b) - c(b-c)(a-b) + c(b-c)(c-a) = a(a-b)(c-a-b+c) - c(b-c)(a-b-c+a) = a(a-b)(3c) - c(b-c)(3a) = 3ac(a-b-b+c) = -9abc

Let Q = \frac {b-c}{a} + \frac {c-a}{b} + \frac {a-b}{c}

Qabc = (b-c)bc + (c-a)ac + (a-b)ab = (b-c)bc + (c-b+b-a)ac + (a-b)ab = (b-c)bc + ac(c-b) + ac(b-a) + (a-b)ab = c(b-c)(b-a) + a(a-b)(b-c) = -(a-b)(b-c)(c-a)

We have: PQabc(b-c)(c-a)(a-b) = (-9abc) * (-(a-b)(b-c)(c-a)) = 9abc(b-c)(c-a)(a-b)

=> PQ = 9 (We have our answer)

After a few trials, it appears that the product is always 9. Observe that $\frac {b-c}{a} ( \frac {b}{c-a} + \frac {c}{a-b} ) = \frac {b-c}{a} \times \frac {ab-b^2+c^2-ca}{(c-a)(a-b)} = \frac {b-c}{a} \times \frac {(b-c)(a-b-c)} {(c-a)(a-b)} = \frac { (b-c)^2 \times 2} {(c-a)(a-b) }$ .

Hence, the expansion is equal to $3 + 2 \times \frac { (b-c)^3 + (c-a)^3 + (a-b)^3} {(b-c)(c-a)(a-b) }$ .

Let $x = b-c, y = c-a$ and $z = a-b$ , so $x+y+z=0$ and the expression is equal to $3 + 2 \times \frac{x^3+y^3+z^3}{xyz}$ . We have $\begin{aligned} 0 &= (x+y+z)^3 \\ &= x^3+y^3+z^3 + 3xy(x+y+z) + 3xz(x+y+z) + 3yz(x+y+z) - 3xyz \\ 3xyz &= x^3+y^3+z^3 \\ \end{aligned}$

Hence the value of the expression is $3 + 2\times 3 = 9$ .

By AM-GM, we have

$[a/(b-c)+b/(c-a)+c/(a-b)]/3 \geq [abc/(b-c)(c-a)(a-b)]^{1/3},$ $[(b-c)/a+(c-a)/b+(a-b)/c]/3\geq [(b-c)(c-a)(a-b)/abc]^{1/3}.$

Multiplying both terms, we get $[a/(b-c)+b/(c-a)+c/(a-b)]/3 \times [(b-c)/a+(c-a)/b+(a-b)/c]/3\geq 1$ . This implies that $[a/(b-c)+b/(c-a)+c/(a-b)] \times [(b-c)/a+(c-a)/b+(a-b)/c] \geq 9$ .