I was just doing an inequality question from a past international competition, but:

a/ I'm not completely certain about my proof

b/ I wonder if there is a simpler and less laborious method of solving it

The question is as follow:

Let $a, b, c$ be positive real numbers such that $abc=1$. Prove that:

$$(a+\frac{1}{b})^2+(b+\frac{1}{c})^2+(c+\frac{1}{a})^2\ge 3(a+b+c+1).$$

I solved it as follows:

$$\begin{aligned}&\quad(a+\frac{1}{b})^2+(b+\frac{1}{c})^2+(c+\frac{1}{a})^2\\
&=a^2+\frac{2a}{b}+\frac{1}{b^2}+b^2+\frac{2b}{c}+\frac{1}{c^2}+c^2+\frac{2c}{a}+\frac{1}{a^2}\\
&=a^2+\frac{a}{b}+\frac{1}{b^2}+b^2+\frac{b}{c}+\frac{1}{c^2}+c^2+\frac{c}{a}+\frac{1}{a^2}+(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\\
&\ge a^2+\frac{a}{b}+\frac{1}{b^2}+b^2+\frac{b}{c}+\frac{1}{c^2}+c^2+\frac{c}{a}+\frac{1}{a^2}+3\quad \text{(AM-GM)}\\
&=a^2+\frac{a}{b}+\frac{ac}{b}+b^2+\frac{b}{c}+\frac{ab}{c}+c^2+\frac{c}{a}+\frac{bc}{a}+3\\
&=a^2+b^2+c^2+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{ac}{b}+\frac{bc}{a}+\frac{ab}{c}+3\\
&=(a^2+\frac{b}{c}+\frac{ac}{b})+(b^2+\frac{c}{a}+\frac{ba}{c})+(c^2+\frac{a}{b}+\frac{bc}{a})\\
&\ge 3\sqrt[3]{a^3}+3\sqrt[3]{b^3}+3\sqrt[3]{c^3}+3\quad\text{(AM-GM)}\\
&=3(a+b+c+1)
\end{aligned}$$

Could you please tell me if my solution is correct and also show me an easier and shorter approach to the question?