Find the maximum value of $x(9\sqrt{1+x^2}​+13\sqrt{1-x^2​})​$

Question

Let $0\le x \le 1$ find the maximum value of $x(9\sqrt{1+x^2}​+13\sqrt{1-x^2​})​$

I try to use am-gm inequality to solve this because it's similar to CMIMC2020 team prob.12 but i don't know how to do next.

Answer 1

Even though I still believe that the calculus approach is the most suited for the problem, consider the somehow magical ansatz I found: \begin{align*}x\cdot\left(9\sqrt{1+x^2}​+13\sqrt{1-x^2​}\right)​&=2\cdot \frac34\cdot 3x\cdot 2\cdot \sqrt{1+x^2}+2\cdot\frac{13}4\cdot x\cdot2\sqrt{1-x^2}\\&\stackrel{AM-GM}{\leqslant}\frac34\cdot\left(9x^2+4\cdot(1+x^2)\right)+\frac{13}4\cdot \left(x^2+4\cdot (1-x^2)\right)\\&=16\end{align*}

Answer 2

You can compute the first derivate of $f(x)$ which is: $$f'(x)=x\left (\frac{9x}{\sqrt{x^2+1}}-\frac{13x}{\sqrt{1-x^2}}\right )+9\sqrt{x^2+1}+13\sqrt{1-x^2}$$ Now you have to study $f'(x)=0$ and so: $$x=\frac{2}{\sqrt{5}}$$ There is another solutions, bit with this we find the local maximum. Now to find the maximim, we compute: $$f\left (\frac{2}{\sqrt{5}}\right)=16$$

Answer 3

By C-S and AM-GM we obtain: $$x\left(9\sqrt{1+x^2}​+13\sqrt{1-x^2​}\right)\leq x\sqrt{(27+13)(3(1+x^2)+13(1-x^2))}=​$$ $$=\sqrt{16\cdot5x^2(8-5x^2)}\leq\sqrt{16\left(\frac{5x^2+8-5x^2}{2}\right)^2}=16.$$ The equality occurs for $x=\frac{2}{\sqrt5},$ which says that $16$ is a maximal value.