Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$ {\left(a+\frac{1}{b}\right)^2}+{\left(b+\frac{1}{c}\right)^2} +{\left(c+\frac{1}{a}\right)^2}≥3(a+b+c+1)$$
My solution:
By Jensen's inequality applied to the convex function $f(x)=x^2$, for $x_1=a+\frac{1}{b} , x_2=b+\frac{1}{c}, x_3=c+\frac{1}{a}$: $$f\left(\frac{\left(a+\frac{1}{b}\right)+\left(b+\frac{1}{c}\right)+\left(c+\frac{1}{a}\right)}{3}\right)≤ \frac{{\left(a+\frac{1}{b}\right)^2}+{\left(b+\frac{1}{c}\right)^2} +{\left(c+\frac{1}{a}\right)^2}}{3}$$ $$\frac{\left(a^2bc+ab^2c+abc^2+bc+ac+ab\right)^2}{a^2b^2c^2}/9≤\frac{{\left(a+\frac{1}{b}\right)^2}+{\left(b+\frac{1}{c}\right)^2} +{\left(c+\frac{1}{a}\right)^2}}{3}$$ If we use the condition $abc=1$: $$\frac{\left(a+b+c+ab+bc+ca\right)^2}{3}≤{\left(a+\frac{1}{b}\right)^2}+{\left(b+\frac{1}{c}\right)^2} +{\left(c+\frac{1}{a}\right)^2}$$ Then we have prove that: $$\frac{\left(a+b+c+ab+bc+ca\right)^2}{3}≥3(a+b+c+1)$$ $$\left(a+b+c+ab+bc+ca\right)^2≥ 9(a+b+c+1)$$ Notice that by A.M-G.M inequality we have $ab+bc+ca≥3$: $$(a+b+c+3)^2≥9(a+b+c+1)$$ If we say $x$ to $a+b+c$ : $$(x+3)^2≥9(x+1)$$ $$x^2+6x+9≥9x+9$$ $$x^2≥3x$$ $$x≥3$$ $$a+b+c≥3$$ This statement is true by A.M-G.M inequality.
If there is another solution, I would be grateful if you could show me :)