But none them resembles me!

$Let\quad 2a=p,\quad 3b=q,\quad 2a+3c=r\\ \\ p,q,r\quad are\quad linearly\quad independent.\\ \\ Now,\cfrac { 3\left( b+c \right) }{ 2a } +\cfrac { 4a+3c }{ 3b } +\cfrac { 12\left( b-c \right) }{ 2a+3c } \\ \\ =\cfrac { q+r-p }{ p } +\cfrac { p+r }{ q } +\cfrac { 4\left( q-r+p \right) }{ r } \\ \\ =\cfrac { p+q+r }{ p } -2+\cfrac { p+q+r }{ q } -1+\cfrac { 4\left( p+q+r \right) }{ r } -8\\ \\ =\cfrac { p+q+r }{ p } +\cfrac { p+q+r }{ q } +\cfrac { 4\left( p+q+r \right) }{ r } -11\\ but\quad AM\ge HM\\ \therefore \cfrac { p+q+2\times \cfrac { r }{ 2 } }{ 4 } \ge \cfrac { 4 }{ \cfrac { 1 }{ p } +\cfrac { 1 }{ q } +2\times \cfrac { 2 }{ r } } \\ \\ \Rightarrow \cfrac { p+q+r }{ p } +\cfrac { p+q+r }{ q } +\cfrac { 4\left( p+q+r \right) }{ r } \ge 16\\ \\ \Rightarrow \cfrac { p+q+r }{ p } +\cfrac { p+q+r }{ q } +\cfrac { 4\left( p+q+r \right) }{ r } -11\ge 5\\ \\ equality\quad hold\quad for\quad p=q=\cfrac { r }{ 2 } \\$

I 'm the head master, i created the problem .Just let you know, there are at least 6 solotions for this problem when i created it. Here is one that easily seen by anyone:

Let $x = \frac{3b + 3c}{2a} + \frac{4a + 3c}{3b} + \frac{12b - 12c}{2a + 3c}$ .

Let $a_{1} = 2a, b_{1} = 3b, c_{1} = 2a + 3c$ .

By AM - GM we have $x = \frac{b_{1} + c_{1} - a_{1}}{a_{1}} + \frac{a_{1} + c_{1}}{b_{1}} + \frac{4[a_{1} + b_{1} - c_{1}]}{c_{1}} = \frac{b_{1}}{a_{1}} + \frac{c_{1}}{a_{1}} + \frac{a_{1}}{b_{1}} + \frac{c_{1}}{b_{1}} + \frac{4a_{1}}{c_{1}} + \frac{4b_{1}}{c_{1}} - 5 \ge 5$ .

Start by assuming that b = c. The motivation for this is simple, because we are trying to minimize the above expression, setting b-c = 0 allows us to get rid of the third term of the expression. Doing so also allows us to now minimize an expression that is in terms of two variables rather than three variables. So, we are trying to minimize 3b/a + 4a/3b + 1. Now, the simplest way to make 3b/a an integer is to allow the variable a to equal the minimum positive multiple of 3, which is 3. Letting a = 3 simplifies and minimizes our expression to b + 4/b + 1. Taking the derivative of this expression and setting it equal to zero allows us to find the critical points of the function to be b = +/- 2, but since a,b,c > 0, we know that b = 2. Now c = 2 since b = c, and it follows that (2,3,3) is the minimum point at which the expression is minimized. Thus, min(f(a,b,c,)) = f(2,3,3) = 5.

More generally, the minimal point happens when b = c and a = 2b/3 = 2c/3! Trivial!

The given expression can be written as :

$[\dfrac{3b}{2a} + \dfrac{2a}{3b} ]+ \dfrac{3c}{2a} + [\dfrac{2a+3c}{3b} + \dfrac{12b}{2a+3c}] - \dfrac{12c}{2a+3c}$

Applying AM GM inequality inside the brackets we get :

$[\dfrac{3b}{2a} + \dfrac{2a}{3b} ]+ \dfrac{3c}{2a} + [\dfrac{2a+3c}{3b} + \dfrac{12b}{2a+3c}] - \dfrac{12c}{2a+3c} \ge 6 + \dfrac{3c}{2a} - \dfrac{12c}{3a+2c}$

The equality occurs when 2a = 3b = 3c

$[\dfrac{3b}{2a} + \dfrac{2a}{3b} ]+ \dfrac{3c}{2a} + [\dfrac{2a+3c}{3b} + \dfrac{12b}{2a+3c}] - \dfrac{12c}{2a+3c} \ge 1+ \dfrac{3c}{2a} + 4 - \dfrac{12c}{2a+3c} + 1$

$[\dfrac{3b}{2a} + \dfrac{2a}{3b} ]+ \dfrac{3c}{2a} + [\dfrac{2a+3c}{3b} + \dfrac{12b}{2a+3c}] - \dfrac{12c}{2a+3c} \ge [\dfrac{2a+3c}{2a} + \dfrac{8a}{2a+3c}] + 1$

Again applying AM GM inequality inside the brackets we get:

$[\dfrac{3b}{2a} + \dfrac{2a}{3b} ]+ \dfrac{3c}{2a} + [\dfrac{2a+3c}{3b} + \dfrac{12b}{2a+3c}] - \dfrac{12c}{2a+3c} \ge 4+1$

Equality occurs when 2a = 3c

Hence the minimum value of the above expression is 5

The inequality is homogenous then we can suppose b=c By AM-GM the result is obvious