Prove the inequality involving exponent and inverse sine

Prove that $$4e^{\operatorname{asin}\left(\frac{1}{6}\right)}-3e^{\operatorname{asin}\left(\frac{1}{9}\right)}-e^{\operatorname{asin}\left(\frac{1}{3}\right)}<0$$

I tried to rewrite it as $3\left(e^{\operatorname{asin}\left(\frac{1}{6}\right)}-e^{\operatorname{asin}\left(\frac{1}{9}\right)}\right)-\left(e^{\operatorname{asin}\left(\frac{1}{3}\right)}-e^{\operatorname{asin}\left(\frac{1}{6}\right)}\right)<0$ but without any success.

Solution 1:

For any $x\in (0,1)$, $$\frac{\mathrm{d} ^2 }{\mathrm{d} x^2}\;e^{\arcsin x}= \frac{(\sqrt{1-x^2}+x)e^{\arcsin x}}{(\sqrt{1-x^2})^3}>0$$

So, $e^{\arcsin x}$ is strictly convex for $x\in (0,1)$. Using Karamata's Majorization Inequality with

$$\left( \frac{1}{6}, \frac{1}{6}, \frac{1}{6}, \frac{1}{6} \right ) \prec \left( \frac{1}{3}, \frac{1}{9}, \frac{1}{9}, \frac{1}{9} \right)$$

We have

\begin{align*}3e^{\arcsin \frac{1}{9}}+e^{\arcsin \frac{1}{3}}&=e^{\arcsin \frac{1}{9}}+e^{\arcsin \frac{1}{9}}+e^{\arcsin \frac{1}{9}}+e^{\arcsin \frac{1}{3}} \\ &>e^{\arcsin \frac{1}{6}}+e^{\arcsin \frac{1}{6}}+e^{\arcsin \frac{1}{6}}+e^{\arcsin \frac{1}{6}} \\ &= 4e^{\arcsin \frac{1}{6}} \end{align*}

easier way by VTand's comment

By Jensen Inequality

\begin{align*}3e^{\arcsin \frac{1}{9}}+e^{\arcsin \frac{1}{3}} &> 4e^{\arcsin (\frac{1/9+1/9+1/9+1/3}{4})}\\ &=4e^{\arcsin \frac{1}{6}} \end{align*}