$(\frac{a}{b}-1)^2+(\frac{b}{a}+1)^2\ge3$
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*}