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

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?

• Try making $b=2-a$ and use derivatives . It works Jul 15 at 6:46

$$b=2-a$$ $$f(a) = \frac{a^2+(2-a)^2}{\sqrt{a^2+1} \sqrt{(2-a)^2+4}}$$ $$f'(a) = \dfrac{2(a^3+6a^2+2a-12)}{\left(\left(2-a\right)^2+4\right)^\frac{3}{2}\left(a^2+1\right)^\frac{3}{2}}$$ 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 $$a=-2+\sqrt{10}$$
• @AnuragA I'm pretty sure my derivative is correct. I've verified it with Desmos and Wolfram|Alpha. Also, plotting the function on Desmos shows that the minimum is at $a = 1.162 \approx \sqrt{10} - 2$ Jul 15 at 7:06