mimimum value of expression $a^2+b^2$ 
If $a,b$ are two non zero real numbers and $ab(a^2-b^2) = a^2+b^2,$ Then $\min(a^2+b^2)$

$\bf{My\; Try::}$ We can write it as $$ab=\frac{a^2+b^2}{a^2-b^2}\Rightarrow a^2b^2=\frac{(a^2+b^2)^2}{(a^2+b^2)^2-4a^2b^2}$$
Now Put $a^2+b^2=u$ and $a^2b^2=v,$ Then expression convert into $$v=\frac{u^2}{u^2-4v}\Rightarrow 4v^2-u^2v+u^2=0$$
For real roots, $\bf{Discriminant \geq 0}$
$$u^4-16u^2\geq 0\Rightarrow u^2(u^2-16)\geq 0$$
So we get $$u^2\geq 16\Rightarrow u\geq 4\Rightarrow x^2+y^2\geq 4,$$
My question is can we solve it any other way (without Trigonometric substution), 
If yes , Then plz explain here, Thanks

If we let $z = a + b i$ then we get 
$$ab = \frac{ z^2 - \bar{z}^2}{4i}$$
$$a^2 - b^2 = \frac{ z^2 + \bar{z}^2}{2}$$
$$a^2 + b^2 = z \bar{z}$$
Thus your original equation can be rewritten as 
$$\frac{Im(z^4)}{4} = \frac{z^4 - \bar{z}^4}{8 i} = \frac{(z^2 - \bar{z}^2)(z^2 + \bar{z}^2)}{8i} = z \bar{z}$$
We have $\frac{\|z\|^4}{4} \geq \frac{Im(z^4)}{4} = \|z\|^2$, so $\|z\|^2 \geq 4$, as desired. This is achieved exactly when $z^4$ is purely imaginary.
