How would you prove $\int^8_0\frac1{\sqrt{x+\frac1{\sqrt{x}}}}dx<4-\frac1{2019}$?

We would like to prove the following inequality.

$$\int^{8}_{0}\frac{1}{\sqrt{x+\frac{1}{\sqrt{x}}}}\,dx<4-\frac{1}{2019}\tag{1}$$

What I've tried is using the AM-GM inequality, $$x+\frac{1}{\sqrt{x}}\ge 2\,\sqrt{x \times\frac{1}{\sqrt{x}}}=2x^{\frac14} $$ then $$\int^{8}_{0}\frac{1}{\sqrt{x+\frac{1}{\sqrt{x}}}}\,dx\le\frac{1}{\sqrt{2}}\int^{8}_{0}\frac{1}{\sqrt[8]{x}}\, dx=4.98\cdots<\color{red}{5}-\frac{1}{2019}.\tag{2}$$How would you prove $(1)$?

It is tempting to upper-bound the integrand by an easier function.

$\color{white}{11110811197115115116117100101110116}$

For $x>0$, one may check that \begin{align} \small{\sqrt {x+\frac{1}{\sqrt{x}}}}&=\sqrt{x}+\frac {1}{2x + \frac 1{2\sqrt{x} +\frac{1}{2x+\frac{1}{\sqrt{x}+\sqrt {x+\frac{1}{\sqrt{x}}}}}}} \tag1 \end{align} then, using \begin{align} \small{\sqrt{x}+\sqrt {x+\frac{1}{\sqrt{x}}}>2\sqrt{x}} \end{align} one gets

$$\int_1^8\frac{dx}{\sqrt {x+\frac{1}{\sqrt{x}}}} <\int_1^8\frac{dx}{\sqrt{x}+\frac {1}{2x + \frac 1{2\sqrt{x} +\frac{1}{2x+\frac{1}{2\sqrt{x}}}}}}. \tag2 $$

By the change of variable $u=2\sqrt{x}$, $du=\frac{dx}{\sqrt{x}}$, setting $\varphi=\frac{1+\sqrt{5}}2$, one has \begin{align} &\int_1^8\frac{dx}{\sqrt{x}+\frac {1}{2x + \frac 1{2\sqrt{x} +\frac{1}{2x+\frac{1}{2\sqrt{x}}}}}} \\\\&=\int_1^{8}\frac{16x^{3}+12x^{3/2}+1}{16x^{3}+20x^{3/2}+5}\:\frac{dx}{\sqrt{x}} \\\\&=\int_2^{4\sqrt{2}}\frac{u^6+6u^3+4}{u^6+10u^3+20}\:du, \qquad u=2^{1/3}5^{1/6}t, \\\\&=4\sqrt{2}-2-2^{4/3}5^{-1/3}\varphi \int_{2^{2/3}5^{-1/6}}^{2^{13/6}5^{-1/6}}\frac{dt}{t^3+\varphi}-\frac{2^{4/3}5^{2/3}}{\varphi}\int_{2^{2/3}5^{-1/6}}^{2^{13/6}5^{-1/6}}\frac{dt}{t^3+\frac1{\varphi}}. \end{align} Then, using the evaluation $$ \int_0^b\frac{dt}{t^3+c^3}=\frac{\sqrt{3}\pi}{18c^2}+\frac{\sqrt{3}}{3c^2}\arctan \left(\frac{2b-c}{\sqrt{3}c}\right)-\frac{1}{6c^2}\ln \left(\frac{(2b-c)^2+3c^2}{4(b+c)^2}\right),\,c>0,\,b>0, $$ one gets

$$ \int_1^8\frac{dx}{\sqrt{x}+\frac {1}{2x + \frac 1{2\sqrt{x} +\frac{1}{2x+\frac{1}{2\sqrt{x}}}}}}=3.3199\color{red}{58928}\cdots. \tag3 $$

In a like manner, for $x>0$, one may check that \begin{align} \small{\sqrt {x+\frac{1}{\sqrt{x}}}}&=\frac{1}{x^{1/4}}+\frac {1}{\frac{2}{x^{5/4}} + \frac 1{\frac{2}{x^{1/4}} +\frac{1}{\frac{2}{x^{5/4}}+\frac{1}{\frac{1}{x^{1/4}}+\sqrt {x+\frac{1}{\sqrt{x}}}}}}} \tag4 \end{align} then, using \begin{align} \small{\frac{1}{x^{1/4}}+\sqrt {x+\frac{1}{\sqrt{x}}}>\frac{2}{x^{1/4}}} \end{align} one gets

$$\int_0^1\frac{dx}{\sqrt {x+\frac{1}{\sqrt{x}}}} <\int_0^1\frac{dx}{\frac{1}{x^{1/4}}+\frac {1}{\frac{2}{x^{5/4}} + \frac 1{\frac{2}{x^{1/4}} +\frac{1}{\frac{2}{x^{5/4}}+\frac{x^{1/4}}2}}}}. \tag5 $$

By the change of variable $u=\sqrt[4]{x}$, $dx=4u^3du$, setting $\varphi=\frac{1+\sqrt{5}}2$, one has

\begin{align} &\int_0^1\frac{dx}{\frac{1}{x^{1/4}}+\frac {1}{\frac{2}{x^{5/4}} + \frac 1{\frac{2}{x^{1/4}} +\frac{1}{\frac{2}{x^{5/4}}+\frac{x^{1/4}}2}}}} \\\\&=\int_0^{1}\frac{x^{3}+12x^{3/2}+16}{5x^{3}+20x^{3/2}+16}\:\sqrt[4]{x}\:dx \\\\&=4\int_0^1\frac{u^{12}+12u^6+16}{5u^{12}+20u^6+16}\:u^4du, \qquad u=2^{1/3}5^{-1/12}t, \\\\&=\frac4{25}+\frac{6}{\sqrt{5}}+2^{-2/3}5^{-1/3}\varphi \int_0^{2^{-1/3}5^{1/12}}\!\!\frac{t^4dt}{t^6+\varphi}-\frac{2^{4/3}5^{-1/3}}{\varphi}\int_0^{2^{-1/3}5^{1/12}}\!\!\frac{t^4dt}{t^6+\frac1{\varphi}}. \end{align}

Then, using the evaluation \begin{align} &6c\int_0^b\frac{t^4dt}{t^6+c^6}\qquad (c>0,\,b>0) \\\\&=\arctan \left(\sqrt{3}+\frac{2b}{c}\right)-\arctan \left(\sqrt{3}-\frac{2b}{c}\right)+2\arctan \left(\frac{b}{c}\right)-\frac{\sqrt{3}}{2}\ln \left(\frac{b^2+\sqrt{3}bc+c^2}{b^2+\sqrt{3}bc+c^2}\right) \end{align} one obtains

$$ \int_0^1\frac{dx}{\frac{1}{x^{1/4}}+\frac {1}{\frac{2}{x^{5/4}} + \frac 1{\frac{2}{x^{1/4}} +\frac{1}{\frac{2}{x^{5/4}}+\frac{x^{1/4}}2}}}}=0.6739\color{red}{21415}\cdots. \tag6 $$

Finally, combining $(2)$, $(3)$, $(5)$ and $(6)$ yields