How to show the minimal value of $\frac{(a+b)^2}{\sqrt{a^2-1}+\sqrt{b^2-4}}$ is 6?

Question

I use mathematical analysis to show the minimal value of $\frac{\left(a + b\right)^{2}}{\,\sqrt{\, a^{2} - 1\,}\, + \,\sqrt{\, b^{2} - 4\,}\,}\left(a > 1\,,\ b >2\right)$ is $6$, but how to show this only by the math knowledge in high school ?

Answer 1

Consider $$f=\frac{(a+b)^2}{\sqrt{a^2-1}+\sqrt{b^2-4}}$$ compute the derivatives and simplify to get $$\frac{\partial f}{\partial a}=\frac{(a+b) \left(2 \sqrt{a^2-1} \sqrt{b^2-4}+a^2-a b-2\right)}{\sqrt{a^2-1} \left(\sqrt{a^2-1}+\sqrt{b^2-4}\right)^2}\tag 1$$ $$\frac{\partial f}{\partial b}=\frac{(a+b) \left(2 \sqrt{a^2-1} \sqrt{b^2-4}-a b+b^2-8\right)}{\sqrt{b^2-4} \left(\sqrt{a^2-1}+\sqrt{b^2-4}\right)^2}\tag 2$$ and set them equal to $0$. So, we are left with $$2 \sqrt{a^2-1} \sqrt{b^2-4}+a^2-a b-2=0 \tag 3$$ $$2 \sqrt{a^2-1} \sqrt{b^2-4}-a b+b^2-8=0\tag 4$$ Subtract $(4)$ from $(3)$ to get $$a^2-b^2+6=0 \implies b=\sqrt{a^2+6}\tag 5$$ Replace in $(3)$ to get $$a^2-\sqrt{a^2+6} a+2 \sqrt{a^2-1} \sqrt{a^2+2}-2=0 \tag 6$$ Now, tedious multiple squaring would lead to $a=\sqrt 2$ and to the result.

I hope and wish that a simpler solution will appear.

Answer 2

Hint

By AM GM

$$\frac {(a+1)+(a-1)}{2}\gt \sqrt {a^2-1}\Rightarrow a\gt \sqrt {a^2-1}$$ And $$\frac {(b+2)+(b-2)}{2}\gt \sqrt {b^2-4}\Rightarrow b\gt \sqrt {b^2-4}$$

Note that there is no equality because the two terms used can never be equal to each other.