<s>[INST] Proving $\frac{a}{b}+\left(\frac{3b+c}{3c+b}\right)^2+\left(\frac{2c+a}{2a+c}\right)^3\ge 3$ for positive $a$, $b$, $c$ 
Let $a,b,c$ be positive real number. Prove that
$$\dfrac{a}{b}+\left(\dfrac{3b+c}{3c+b}\right)^2+\left(\dfrac{2c+a}{2a+c}\right)^3\ge 3$$

It is not a symmetric inequality. I know that the equality occur when $a=b=c$ but It's hard to know  how to approach it.
First, let $x=\dfrac{a}{b},y=\dfrac{b}{c},z=\dfrac{c}{a}$, then $xyz=1$ and then inequality becomes
$$x+\left(\dfrac{3y+1}{y+3}\right)^2+\left(\dfrac{2z+1}{z+2}\right)^3\ge 3 \tag1$$
My direction is use UCT method but
$$\left(\dfrac{3y+1}{y+3}\right)^2\ge y\quad\Leftrightarrow\quad y\le 1 \tag2$$ and similar to $$\left(\dfrac{2z+1}{z+2}\right)^3\ge y \tag3$$ and I don't have any information about $y,z$.
I wonder if it was the right way to solve it! Please give me some hint! Thanks!
 [/INST] Using the AM-GM inequality,
$$\left(\frac{3b+c}{3c+b}\right)^2+1\ge2\left(\frac{3b+c}{3c+b}\right) $$
$$\left(\frac{2c+a}{2a+c}\right)^3+1+1\ge3\left(\frac{2c+a}{2a+c}\right)\;\;\;\;\;$$
$$\therefore\;\;\frac{a}{b}+\left(\frac{3b+c}{3c+b}\right)^2+\left(\frac{2c+a}{2a+c}\right)^3+3\ge \frac{a}{b}+2\left(\frac{3b+c}{3c+b}\right)+3\left(\frac{2c+a}{2a+c}\right)\;\;\;$$
It remains to prove:$$\frac{a}{b}+2\left(\frac{3b+c}{3c+b}\right)+3\left(\frac{2c+a}{2a+c}\right)\ge 6 \;\Longleftrightarrow \frac{a-b}{b}+4\left(\frac{b-c}{3c+b}\right)+3\left(\frac{c-a}{2a+c}\right)\ge0$$
Expanding the left hand side and completing the square leads to
$$6c(a-b)^2+2b(a-c)^2+3a(b-c)^2\ge0$$
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