A Small Touch Of Substitution

Let's try to make the expression symmetric first, that is, try to make the values of $a,b$ and $c$ interexchangeable. With a little fiddling (hard to explain this part), let $a = A, b = 2B, c = C$ , the expression becomes

$\begin{aligned} \dfrac{4C}{2A+2B} + \dfrac{4A}{2B+2C}+ \dfrac{2B}{A+C} &=& \dfrac{2C}{A+B} + \dfrac{2A}{B+C}+ \dfrac{2B}{A+C} \\ &=& 2 \left( \dfrac C{A+B} + \dfrac A{B+C} + \dfrac B{A+C} \right) \end{aligned}$

This is a direct application of Titu's lemma , or more specifically Nesbitt's inequality (as shown in the fourth example):

For $a,b,c\in\mathbb{R^{+}}$ , $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \dfrac{3}{2}.$

Thus, $\dfrac{4C}{2A+2B} + \dfrac{4A}{2B+2C}+ \dfrac{2B}{A+C} \geq 2 \cdot \dfrac32 = 3$ .

So the minimum value of the expression in question is $\boxed3$ and it occurs when $A=B=C$ or $2a=b=2c \Rightarrow a = \dfrac14 , b = \dfrac12, c = \dfrac14$ .

Call the expresion $A$ , now we see $A+6=\frac{4c+4a+2b}{2a+b}+\frac{2b+4a+4c}{2c+b}+\frac{2a+2c+b}{c+a}$ $\Leftrightarrow A+6=(2a+2c+b)\bigg(\frac{1}{a+\frac{b}{2}}+\frac{1}{c+\frac{b}{2}}+\frac{1}{c+a}\bigg)$ By Cauchy-Schwarz inequality , we easily prove that $A+6\geq 9$ $\Leftrightarrow A\geq 3$ The equality holds when $b=2a=2c$

$\frac{4c}{2a+b}+\frac{4a}{b+2c}+\frac{b}{c+a}=\frac{4c^2}{2ac+bc}+\frac{4a^2}{ab+2ac}+\frac{b^2}{bc+ab} \geq \frac{(2c+2a+b)^2}{4ac+2bc+2ab}$ Let $2c=x,2a=y,b=z$ .Then the expression becomes: $\frac{4c}{2a+b}+\frac{4a}{b+2c}+\frac{b}{c+a}\geq \frac{(x+y+z)^2}{xy+xz+yz}$ Notice that: $\begin{aligned} x^2+y^2+z^2 & \geq xy+xz+yz\\ \implies (x+y+z)^2 & \geq 3(xy+xz+yz)\\ \implies \dfrac{(x+y+z)^2}{xy+xz+yz} & \geq 3 \end{aligned}$ Hence: $\frac{4c}{2a+b}+\frac{4a}{b+2c}+\frac{b}{c+a}\geq \boxed{3}$ Equality occurs when $2c=2a=b$ .