Mfrhtyj

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, $$x+frac{1}{sqrt{x}}ge 2,sqrt{x timesfrac{1}{sqrt{x}}}=2x^{frac14} $$ then $$int^{8}_{0}frac{1}{sqrt{x+frac{1}{sqrt{x}}}},dxlefrac{1}{sqrt{2}}int^{8}_{0}frac{1}{sqrt[8]{x}}, dx=4.98cdots<color{red}{5}-frac{1}{2019}.tag{2}$$How would you prove $(1)$?

$color{white}{11110811197115115116117100101110116}$

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

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

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

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

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

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

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

begin{align}
&int_0^1frac{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
\\&=4int_0^1frac{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}
&6cint_0^bfrac{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)+2arctan 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

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

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

$$int_0^8frac{dx}{sqrt {x+frac{1}{sqrt{x}}}} <color{red}{3.99}38803cdots<color{red}{4}-frac1{2019}=color{red}{3.99}950cdots tag7
$$

as desired.