Maximum Value

For this problem we can use the Geometric - Arithmetic - Quadratic mean inequality.

First, we know that

$\sqrt [ 3 ]{ abc } \le \frac { a+b+c }{ 3 }$ which is $\sqrt [ 3 ]{ abc } \le \frac { 1 }{ 3 }$ involving $abc\le \frac { 1 }{ 27 }$ . So, we must take the maximum value of $abc$ which is $\frac { 1 }{ 27 }$ so $2abc=\frac { 2 }{ 27 }$ .

Then, we know

${ (a+b+c) }^{ 2 }=({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 })+2(ab+bc+ca)=1$ . Now, $\frac { 1 }{ 3 } =\frac { a+b+c }{ 3 } \le \sqrt { \frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ 3 } }$ therefore, $\frac { 1 }{ 3 } \le { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }$ . Then, $2(ab+bc+ca)\le \frac { 2 }{ 3 }$ which is $ab+bc+ca\le \frac { 1 }{ 3 }$ , and as we want to take the maximum value $ab+bc+ca=\frac { 1 }{ 3 }$ .

So now we replace the values in the first equation $ab+bc+ca-2abc$ equivalent to $\frac { 1 }{ 3 } -\frac { 2 }{ 27 } =\frac { 7 }{ 27 } =\frac { m }{ n }$ so m+n=27+7=\boxed { 34 } .

Let $ab+bc+ca-2abc=X$ . By AM-GM inequality: $(1-2a)(1-2b)(1-2c) \le \left(\dfrac{3-2(a+b+c)}{3}\right)^3=\dfrac{1}{27}$ Expanding the left side gives us: $1-2(a+b+c)+4(ab+bc+ca-2abc)=4X-1$ Hence: $4X-1 \le \dfrac{1}{27} ~~ \Longrightarrow ~ X \le \boxed{\dfrac{7}{27}}$ with equality iff $1-2a=1-2b=1-2c~\Longrightarrow a=b=c=\dfrac{1}{3}$ .

This problem can be answered by the postulate: "if the sum is constant, the product is maximum when all of terms are equal"