If $\alpha = 2\arctan(2\sqrt{2}-1)$ and $\beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;,$ Then prove that $\alpha>\beta$ 
If $\alpha = 2\arctan(2\sqrt{2}-1)$ and $\displaystyle \beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;,$ Then prove that $\alpha>\beta$

$\bf{My\; Try::}$ Given $$ \alpha = \arctan \bigg(\frac{2(2\sqrt{2}-1)}{1-(2\sqrt{2}-1)^2}\bigg)=\arctan \bigg(\frac{(2\sqrt{2}-1)}{2\sqrt{2}-4}\bigg)$$
and $$\beta = \arcsin\bigg(1-\frac{4}{27}\bigg)+\arcsin\bigg(\frac{3}{5}\bigg) = \arcsin\bigg(\frac{23}{27}\bigg)+\arcsin\bigg(\frac{3}{5}\bigg)$$
So $$\arcsin\bigg[\frac{23}{27}\cdot \frac{4}{5}+\frac{3}{5}\cdot \frac{10\sqrt{2}}{23}\bigg]$$
Now how can i solve it, Help required, Thanks
 A: HINT:
As $\arcsin\dfrac13<\arcsin\dfrac12=\dfrac\pi6$
$3\arcsin\dfrac13=\arcsin\dfrac{23}{27}=\arctan\dfrac{23}{10\sqrt2}$
$\arcsin\dfrac35=\arctan\dfrac34$
As $\dfrac{23}{10\sqrt2}\cdot\dfrac34>1,$
using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,
$$\arctan\dfrac{23}{10\sqrt2}+\arctan\dfrac34=\pi+\arctan\dfrac{\dfrac{23}{10\sqrt2}+\dfrac34}{1-\dfrac{23}{10\sqrt2}\cdot\dfrac34}$$
As $2\sqrt2-1>1$
$$2\arctan(2\sqrt{2}-1)=\pi+\arctan ?$$
Now for $\dfrac\pi2>a>b>-\dfrac\pi2,\arctan a>\arctan b $
A: On the one hand we have
$$\tan{\alpha\over2}=2\sqrt{2}-1>\sqrt{3}=\tan{\pi\over3}\ ,$$
hence ${\alpha\over2}>{\pi\over3}$, or $$\alpha>{2\pi\over3}\ .$$
On the other hand the convexity of $$t\mapsto\arcsin t\qquad(0\leq t\leq{\pi\over2})$$ implies
$$\arcsin{1\over3}<{2\over3}\arcsin{1\over2}={2\over3}\>{\pi\over6}={\pi\over9}\ .$$
Furthermore $\arcsin{3\over5}<\arcsin{1\over\sqrt{2}}={\pi\over4}$. It follows that
$$\beta<3{\pi\over9}+{\pi\over4}={7\pi\over12}<\alpha\ .$$
A: \begin{align}
\beta&=\sqrt{10}\left(\frac{3}{\sqrt{10}}\arcsin\frac13+\frac{1}{\sqrt{10}}\arcsin\frac35\right)\\
&<\sqrt{10}\arcsin\left(\frac{4\sqrt{10}}{25}\right)
\end{align}
where I use the convexity of $\arcsin$. On the other hand using the definition of $\arctan$ we obtain $$\arcsin\left(\frac{4\sqrt{10}}{25}\right)=\arctan\left(\frac{4\sqrt{2}}{\sqrt{93}}\right).$$
Now further note that $\sqrt{10}<4$ hence we should check wether $$2\arctan\left(\frac{2\sqrt{2}}{\sqrt{93}}\right)<\arctan(2\sqrt2-1)$$
We hence (with $\tan2x=\frac{2\tan x}{1-\tan^2x}$) should prove that $$\frac{8\sqrt{186}}{61}<2\sqrt2-1.$$
