Question about Sui Zhen Lin's proof for inequality $\sqrt{\frac{a^2}{9b^2-8b+4}}+\sqrt{\frac{4b}{a+4}}\leq 1$ with positive numbers $a,b$ so $a+b=1$ given two positive numbers $a, b$ so that $a+ b= 1$

Sui Zhen Lin ; @szl6208 gave a very beautiful proof for the following inequality
$$\sqrt{\frac{a^{2}}{9b^{2}- 8b+ 4}}+ \sqrt{\frac{4b}{a+ 4}}\leq 1$$
Source: AoPS/@szl6208_ on.AoPS

Proof. We have
$$\sqrt{\frac{a^{2}}{9b^{2}- 8b+ 4}}\leq\frac{\sqrt{a\left ( 9b^{2}- 8b+ 4 \right )\cdot a\left ( a+ 1 \right )^{2}}}{\left ( a+ 1 \right )\left ( 9b^{2}- 8b+ 4 \right )}\leq\frac{a\left ( 9b^{2}- 8b+ 4 \right )+ a\left ( a+ 1 \right )^{2}}{2\left ( a+ 1 \right )\left ( 9b^{2}- 8b+ 4 \right )}$$
$$\sqrt{\frac{4b}{a+ 4}}\leq\frac{\sqrt{4b\left ( a+ 4 \right )\cdot\left ( 2b+ 2 \right )^{2}}}{\left ( a+ 4 \right )\left ( 2b+ 2 \right )}\leq\frac{4b\left ( a+ 4 \right )+ \left ( 2b+ 2 \right )^{2}}{2\left ( a+ 4 \right )\left ( 2b+ 2 \right )}$$
Hence, we need to prove that
$$\frac{a\left ( 9b^{2}- 8b+ 4 \right )+ a\left ( a+ 1 \right )^{2}}{2\left ( a+ 1 \right )\left ( 9b^{2}- 8b+ 4 \right )}+ \frac{4b\left ( a+ 4 \right )+ \left ( 2b+ 2 \right )^{2}}{2\left ( a+ 4 \right )\left ( 2b+ 2 \right )}\leq 1$$
$$\Leftrightarrow RHS- LHS= \frac{a\left ( ab+ 11a+ 1 \right )\left ( a- b \right )^{2}}{\left ( a+ 1 \right )\left ( 9b^{2}- 8b+ 4 \right )\left ( a+ 4 \right )\left ( b+ 1 \right )}\geq 0$$
wait a minute, actually
$$RHS- LHS- \frac{a\left ( ab+ 11a+ 1 \right )\left ( a- b \right )^{2}}{\left ( a+ 1 \right )\left ( 9b^{2}- 8b+ 4 \right )\left ( a+ 4 \right )\left ( b+ 1 \right )}= \left ( a+ b- 1 \right )f\left ( a, b \right )= 0$$
but how.. why is that so perfect ? I think there maybe are solutions constructed by substitutions like $g\left ( a \right )- g\left ( 1- b \right )\!, g\left ( 2a \right )- g\left ( a+ 1- b \right )\!= \left ( a+ b- 1 \right )f\left ( x \right ).$ What's the best perfect substitution here ? Someone teaching me, huh ?

Using algebra.
From the constraint $b=1-a$. So, we face
$$f(a)=\sqrt{\frac{a^2}{9 a^2-10 a+5}}+2 \sqrt{\frac{1-a}{a+4}}$$
$$f'(a)=\frac{5-5 a}{(9 a^2-10 a+5)^{3/2}}-\frac{5}{\sqrt{1-a} (a+4)^{3/2}}$$
Square and factor; the numerator is just
$$(2 a-1) (5 a-1) \left(a^2-a+1\right) \left(73 a^2-118 a+61\right)$$ So, the derivative only cancels for $a=\frac 15$ and $a=\frac 12$. The second derivative test shows the the first is a minimum and the second a maximum. So, we have
$$f\left(\frac{1}{5}\right)=\frac 32\sqrt{\frac{3}{7}} \qquad \text{and} \qquad f\left(\frac{1}{2}\right)=1$$
Around $a=\frac 12$, we have
$$f(a)=1-\frac{20}{27} \left(a-\frac{1}{2}\right)^2-\frac{1480}{729}
   \left(a-\frac{1}{2}\right)^3+O\left(\left(a-\frac{1}{2}\right)^4\right)$$
