if $a+b=2$, finding minimum value of $\frac{a^2+b^2}{\sqrt{a^2+1} \sqrt{b^2+4}}$

Question

if $a+b=2$, Finding minimum value of $\frac{a^2+b^2}{\sqrt{a^2+1} \sqrt{b^2+4}}$.

I only thought of the Lagrangian multiplier method, it’s a lot of calculation, I can’t do it.
$\frac{a^2+b^2}{\sqrt{a^2+1} \sqrt{b^2+4}}+\lambda (a+b-2)$
(1) $-\frac{a \left(a^2+b^2\right)}{\left(a^2+1\right)^{3/2} \sqrt{b^2+4}}+\frac{2 a}{\sqrt{a^2+1} \sqrt{b^2+4}}+\lambda =0$
(2) $-\frac{b \left(a^2+b^2\right)}{\sqrt{a^2+1} \left(b^2+4\right)^{3/2}}+\frac{2 b}{\sqrt{a^2+1} \sqrt{b^2+4}}+\lambda =0$

Is there a simpler way?

Answer 1

$$ b=2-a$$ $$ f(a) = \frac{a^2+(2-a)^2}{\sqrt{a^2+1} \sqrt{(2-a)^2+4}} $$ $$ f'(a) = \dfrac{2(a^3+6a^2+2a-12)}{\left(\left(2-a\right)^2+4\right)^\frac{3}{2}\left(a^2+1\right)^\frac{3}{2}}$$ The denominator of the derivative is always positive. It is sufficient to analyse the numerator. Observe that $a=-2$ is a root of the cubic in the numerator. It can be factorised as $$ a^3+6a^2+2a-12 = (a+2)(a^2+4a-6)$$ The critical points of this function are thus $-2, -2 \pm \sqrt{10} $
It's easy to see that $a=-2$ is the local maxima and $a= -2 \pm \sqrt{10} $ are the local minima. Now compare the value of $f(a)$ at these two points to see which one is smaller. The minimum of $f(a)$ comes out to be at $a=-2+\sqrt{10}$