Cubes to square root

Without losing generality we assume that the maximum can be achieved when $x^3=a$ , $y^3=b$ , by AM-GM $x^3+5a\geq 6\sqrt[6]{x^3a^5}=6\sqrt{x}\sqrt[6]{a^5}$ $y^3+5b\geq 6\sqrt{y}\sqrt[6]{b^5}$ Now we need to have $2\sqrt[6]{a^5}=\sqrt[6]{b^5}$ $\therefore 1+5(a+b)\geq x^3+y^3+5(a+b)\geq 6\sqrt{x}\sqrt[6]{a^5}+6\sqrt{y}\sqrt[6]{b^5}$ $\Leftrightarrow 1+5(1+2\sqrt[5]{2})a\geq 6\sqrt[6]{a^5}(\sqrt{x}+2\sqrt{y})$ $\Leftrightarrow \sqrt{x}+2\sqrt{y}\leq\frac{1+5(1+2\sqrt[5]{2})a}{6\sqrt[6]{a^5}}=\frac{1}{6\sqrt[6]{a^5}}+\frac{5}{6}(1+2\sqrt[5]{2})\sqrt[6]{a}$ Now, $a,b$ must satisfy $\begin{cases}2\sqrt[5]{2}a=b\\a+b=1\end{cases}$ $\Leftrightarrow\begin{cases}a=\frac{1}{1+2\sqrt[5]{2}}\\b=\frac{\sqrt{2\sqrt[5]{2}}}{1+2\sqrt[5]{2}}\end{cases}$ $\therefore \sqrt{x}+2\sqrt{y}\leq \frac{1}{6\sqrt[6]{\frac{1}{(1+2\sqrt[5]{2})^5}}}+\frac{5}{6}\sqrt[6]{(1+2\sqrt[5]{2})^5}$ $\Leftrightarrow \sqrt{x}+2\sqrt{y}\leq \sqrt[6]{(1+2\sqrt[5]{2})^5}\approx 2.70$ The equality holds when $(x,y)=\bigg(\frac{1}{\sqrt[3]{1+2\sqrt[5]{2}}};\sqrt[3]{\frac{2\sqrt[5]{2}}{1+2\sqrt[5]{2}}}\bigg)$

$Let~x=1-n~and~y=n,~~since~we~want~maximum.\\ So~\Huge f=\sqrt x+2*\sqrt y=\sqrt[6] {1-n}+2*\sqrt[6] {n}.\\ Using~this~in~the~keisan~calculator,\\ n=0,.1,11\\ We~get:-\\$
n ..... ...... f
0 .... ...... 1
0.1 ...... 2.34518
0.2...... 2.49294
0.3 ...... 2.57866
0.4 ...... 2.63513
0.5 ...... 2.6727
0.6 ...... 2.69515
$\color{#D61F06}{0.7~~~~~~ 2.70276}$
0.8 ...... 2.69171
0.9 ...... 2.64648
1 ........ 2
To refine the value, next I used,
n=.6,.02,21
I got the best as n=7.0 .....f=2.70276
Further n=.68,.001,20
I got the best at n=6.97 .........f= 2.702771453
$So~\sqrt x+2*\sqrt y=\Large \color{#D61F06}{2.702771453}.$