How find this range of the fucntion $f(a,b)=\left(\frac{1}{a}+\frac{1}{b}+1\right)(a-3b+15)$ let $a,b>0$,and such $a+2b=3$,find  range of the follow function 
$$f(a,b)=\left(\dfrac{1}{a}+\dfrac{1}{b}+1\right)(a-3b+15)$$
My idea: since
\begin{align*}f(a,b)&=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{a+2b}{3}\right)(a-3b+5(a+2b))
=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{a+2b}{3}\right)(6a+7b)\\
&=6+\dfrac{7b}{a}+\dfrac{6a}{b}+7+\dfrac{1}{3}(6a^2+19ab+14b^2)
\end{align*}
Follow I want use AM-GM inequality,But I can't .Thank you
 A: As mentioned above, as $b \to 0_+$, we have $f(a, b) \to \infty$, so there is no maximum.  For finding the minimum, it seems expressing it as a function of one variable (using the constraint) and calculus seems straightforward though messy.  If you really must use inequalities, here is an approach.
$f(1, 1) = 39$, so the minimum cannot be higher than this.   By rewriting the inequality and Cauchy-Schwarz, we have
$$\left(\frac{1}{a}+\frac{1}{b}+1\right)(a-3b+15) = \left(\frac{1}{a}+\frac{1}{b}+1\right)(5a+5b+3)\ge \left(\sqrt5+\sqrt5+\sqrt3\right)^2 \approx  38.49$$
As the equality cannot be achieved, this value is never achieved, but it shows that the minimum must be higher than this.  So this gives us for the range that $f \in [m, \infty)$, where $38.49 < m \le 39$.
If the bound isn't sufficient and you need the exact value, note that we can write
$$a-3b+15 = a-3b+15-3k + k(a+2b) = (k+1)a+(2k-3)b+15-3k$$
So for any real $k \in [\frac32, 5]$, we can have by CS inequality:
$$f(a, b) = \left(\frac{1}{a}+\frac{1}{b}+1\right)\left((k+1)a+(2k-3)b+(15-3k \right)\ge \left(\sqrt{k+1}+\sqrt{2k-3}+\sqrt{15-3k}\right)^2 $$
We can find the minimum only if we can have equality condition for some $k$, i.e. we must have a solution for allowable $a, b, k$ s.t.
$$(k+1)a^2= (2k-3)b^2= 15-3k$$
While it looks terrible to try solving analytically, numerically, $a \approx 0.967588, b \approx 1.01621, k \approx 3.57291$ works, giving a minimum of $f\ge m \approx 38.9836$
