Find the maximum of E_a when the entropy is at its maximum

Homework Statement

Show that $\Gamma_T$ is maximum at $E_a = \frac{N_aE}{N_a+N_b}$

Homework Equations

The expression for $\Gamma(E)$ when $N\gg 1$
$$\Gamma_T = C_aC_b exp \left( -\frac{E_a^2}{2N_a\mu_B^2h^2} \right) exp \left[ - \frac{(E-E_a)^2}{2N_b\mu_B^2H^2} \right]$$
where $C_a$ and $C_b$ are normalization constants that depends on $N_a$ and $N_b$
In terms of the entropy:
\begin{aligned} \frac{1}{k}S_T = \frac{1}{k}(S_a+S_b) &= ln \Gamma_T \\ &= ln(C_aC_b)-\frac{1}{2(\mu_BH)^2} \left[ \frac{E_a^2}{N_a} + \frac{(E-E_a)^2}{N_b} \right]\end{aligned}

The Attempt at a Solution

Try to find the maximum of the entropy using the quadratic terms in the square bracket:
\begin{aligned} \frac{E_a^2}{N_a} + \frac{(E-E_a)^2}{N_b} &= \frac{(N_a+N_b)E_a^2}{N_aN_b} - \frac{2EE_a}{N_b} + \frac{E^2}{N_b} \\ &= \frac{N_a+N_b}{N_aN_b} \left[ E_a^2 - \frac{2N_aN_bEE_a}{N_b(N_a+N_b)} \right] + \frac{E^2}{N_b} \\ &= \frac{(N_a+N_b)}{N_aN_b} \left[ E_a - \frac{N_aE}{N_a+N_b} \right]^2 +\frac{E^2}{N_b} - \frac{N_a^2E^2}{(N_a+N_b)^2} \end{aligned}
And the solution manual reduce this equation to
$$\frac{(N_a+N_b)}{N_aN_b} \left[ E_a - \frac{N_aE}{N_a+N_b} \right]^2 + \frac{E^2}{N_a+N_b}$$
I can't get the equality of
$$\frac{E^2}{N_b} - \frac{N_a^2E^2}{(N_a+N_b)^2}=\frac{E^2}{N_a+N_b}$$
With $$\frac{E^2[N_a^2+2N_aN_b+N_b^2-N_bN_a^2]}{N_b(N_a+N_b)^2}=\frac{E^2(N_a+N_b)N_b}{N_b(N_a+N_b)^2}$$ I just cant get both side of the equation to be equal...

kuruman
Sorry, what I meant was I can't factor the equation $$N_a^2 + 2N_aN_b + N_b^2 -N_bN_a^2$$ to $$(N_a+N_b)N_b$$