To prove the inequality of positive rational numbers Show that: $$ \left(\frac{a+b}{a+b+c}\right)^{c} \left(\frac{b+c}{a+b+c}\right)^{a} \left(\frac{a+c}{a+b+c}\right)^{b}< \left(\frac{2}{3}\right)^{a+b+c} ,a\ne b\ne c$$
PS: I am supposed to use weighted AM > GM > HM for this problem.
I tried by expressing $$\frac{a+b}{a+b+c} = 1- \frac{c}{a+b+c} $$ but I cannot get required expression.

Edit: Fixed algebra error.
Consider the weighted AM-GM with weights: $\frac{a}{a+b+c}, \frac{b}{a+b+c}, \frac{c}{a+b+c}$. We attain:
$$(\frac{a(b+c) + b(a+c) + c(a+b)}{a+b+c})^{a+b+c} > (a+b)^c (b+c)^a (a+c)^b.$$
And thus:
$$(\frac{a(b+c) + b(a+c) + c(a+b)}{(a+b+c)^2})^{a+b+c} > (\frac{a+b}{a+b+c})^c (\frac{b+c}{a+b+c})^a (\frac{a+c}{a+b+c})^b.$$
So it's sufficient to show $\frac{2}{3} > (\frac{a(b+c) + b(a+c) + c(a+b)}{(a+b+c)^2})$. Note that
$$\frac{a(b+c) + b(a+c) + c(a+b)}{(a+b+c)^2} = \frac{(a + b+c)^2 - (a^2 + b^2 + c^2)}{(a+b+c)^2}.$$
But by AM-QM: $a^2 + b^2 + c^2 \geq \frac{(a+b+c)^2}{3}$, so the RHS is less than or equal to
$$\frac{(a+b+c)^2 - \frac{(a+b+c)^2}{3}}{(a+b+c)^2} = \frac{2}{3}.$$
