$(\frac{a}{b}-1)^2+(\frac{b}{a}+1)^2\ge3$ I tried expanding the parenthesis and then grouping them back somehow but i couldn't get anywhere.
I also tried to apply known inequalities with no success. Can someone help me out?

Substitute $x=\frac{a}{b}$.
\begin{align*}
\left(\frac{a}{b}-1\right)^2+\left(\frac{b}{a}+1\right)^2 - 3 &= (x-1)^2+\left(1+\frac{1}{x}\right)^2-3 \\
&= \frac{x^2(x^2-2x+1)+(x^2+2x+1)-3x^2}{x^2} \\
&= \frac{x^4-2x^3-x^2+2x+1}{x^2} \\
&= \frac{(x^2-x-1)^2}{x^2} \\
&\geq 0.
\end{align*}
