Greater of the angles $\alpha=2\tan^{-1}(2\sqrt{2}-1)$, $\beta=3\sin^{-1}\frac{1}{3}+\sin^{-1}\frac{3}{5}$ 
Find the greater of the two angles $\alpha=2\tan^{-1}(2\sqrt{2}-1)$ and $\beta=3\sin^{-1}\dfrac{1}{3}+\sin^{-1}\dfrac{3}{5}$

My Attempt
$$
\alpha=2\tan^{-1}(2\sqrt{2}-1)=2\tan^{-1}(1.82)>2\tan^{-1}\sqrt{3}=2.\frac{\pi}{3}\\
\implies \boxed{\alpha>\frac{2\pi}{3}=0.67\pi}\\
\beta=3\sin^{-1}\dfrac{1}{3}+\sin^{-1}\dfrac{3}{5}=3\sin^{-1}(0.33)+\sin^{-1}(0.6)\\
=\sin^{-1}(0.8)+\sin^{-1}(0.6)<\sin^{-1}(1)+\sin^{-1}(0.7=1/\sqrt{2})\\
\boxed{\beta<\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}=0.75\pi}
$$
I am stuck with the above result which does not seem to give enough information to decide which one is greater, wht's the easiest way to  solve this ?

Note that $\sqrt{2}\approx1.414$.
$(2\sqrt{2}-1)^2=8-4\sqrt{2}+1=9-4\sqrt{2}>3\Leftrightarrow6>4\sqrt{2}\Leftrightarrow $ after raising to the second power $\Leftrightarrow 36>32$ which is true and $2\sqrt{2}-1>\sqrt{3}$
We know that $\tan\dfrac{\pi}{3}=\sqrt{3}$, we get that $(2\sqrt{2}-1)>\tan\dfrac{\pi}{3}$, then $2\tan^{-1}(2\sqrt{2}-1)>\dfrac{2\pi}{3}$, because the inverse of the tangent is an increasing function.
So, $\alpha>\dfrac{2\pi}{3}$
Note that $\sin3\theta=3\sin\theta-4\sin^2\theta$, then $3\theta=\sin^{-1}[3\sin\theta-4\sin^3\theta]$.
We have that $\theta=\sin^{-1}\dfrac13$ and $\sin\theta=\dfrac13$
So, $$\beta=3\sin^{-1}\dfrac13+\sin^{-1}\dfrac35$$
$$=\sin^{-1}\left[3\left(\dfrac13\right)-4\left(\dfrac13\right)^3\right]+\sin^{-1}\dfrac35$$
$$=\sin^{-1}\dfrac{23}{27}+\sin^{-1}\dfrac{3}{5}$$
Note that $\dfrac{23}{27}<\dfrac{\sqrt{3}}{2}$ and $\dfrac35<\dfrac{\sqrt{3}}{2}$, because $(23\cdot2)^2<(27\sqrt{3})^2$ and $(3\cdot2)^2<(5\sqrt{3})^2$, which is equivalent to $2116<2187$ and $36<75$ which is true.
So, $\sin^{-1}\dfrac{23}{27}+\sin^{-1}\dfrac35<\sin^{-1}\dfrac{\sqrt{3}}{2}+\sin^{-1}\dfrac{\sqrt{3}}{2}=\dfrac{\pi}{3}+\dfrac{\pi}{3}=\dfrac{2\pi}{3}$, because the inverse of the sine function is strictly increasing.
Therefore, $$\alpha>\dfrac{2\pi}{3},\ \beta<\dfrac{2\pi}{3}$$
$$\mbox{So, }\alpha>\beta$$
