Prove that $\sqrt {2a+{\frac {36-8a}{a+9}}+2\sqrt {{a}^{2}+3}}+\sqrt {{ \frac {36-8a}{a+9}}}\leqslant 3+\sqrt{3}$ For $0\leqslant a \leqslant 3.$ Prove that:
$$f(a)=\sqrt {2a+{\frac {36-8a}{a+9}}+2\sqrt {{a}^{2}+3}}+\sqrt {{
\frac {36-8a}{a+9}}}\leqslant 3+\sqrt{3}$$

The text for LHS:  sqrt(2a+(36-8a)/(a+9)+2sqrt(a^2+3))+sqrt((36-8a)/(a+9))

I found that the equality holds when $a=3$ or $a=1.$ I also found that when $3\ge a\ge 1$ then $f'(a)\ge 0$ but I can not prove it and I don't know what to do in the opposite case. Helps me!
 A: Consider that you look for the extrema of function $$f(a)=\sqrt {2a+{\frac {36-8a}{a+9}}+2\sqrt {{a}^{2}+3}}+\sqrt {{ \frac {36-8a}{a+9}}}$$ Its ugly first derivative is almost $0$ very close to $a=1$
$$f'(1)=\frac{12}{5 \sqrt{55}}-\frac{27}{10 \sqrt{70}}\sim  0.000904209$$ Similarly
$$f''(1)=\frac{4161}{4400 \sqrt{55}}+\frac{27}{1400 \sqrt{70}}\sim 0.129821 >0$$ confirms that it is a minimum. We also have
$$f(1)=\frac{1}{5} \left(2 \sqrt{55}+\sqrt{70}\right)\sim  4.63980$$
Now, you "just" need to prove that, in the range, $f'(a)$ does not cancel a second time.
Edit
If we make a series expansion around $a=1$, the minimum is
$$a_{min} \sim \frac{27 \left(5687 \sqrt{14}-3871 \sqrt{11}\right)}{67963 \sqrt{11}+1089 \sqrt{14}}\approx 0.993035$$
$$f(a_{min} )\approx 4.639796308297$$ while a full optimization gives
$$a_{min}\approx 0.993066 \qquad \text{and} \qquad f(a_{min} )\approx 4.639796308235$$
In the same spirit, we could show that the first derivative goes though a maximum close to $a=2$ (exactly at $a\approx 2.06292$) and an expansion around $a=2$ shows that it would cancel again around $a\approx 3.23940 >3$ (the exact value would be $a\approx3.24253$).
