How to find the maximum of the value $\frac{\sin{x}+1}{\sqrt{3+2\cos{x}+\sin{x}}}$ Find the maximun of the value
$$f(x)=\dfrac{\sin{x}+1}{\sqrt{3+2\cos{x}+\sin{x}}}$$
I use wolframpha this found this maximum is $\dfrac{4\sqrt{2}}{5}$,But How to prove  and how to find this value?(without derivative)
idea 1
let $\tan{\dfrac{x}{2}}=t$,then we have
$$f=\dfrac{(t+1)^2}{\sqrt{t^4+2t^3+6t^2+2t+5}}$$
Therefore,it suffices to prove that
$$\dfrac{(t+1)^4}{t^4+2t^3+6t^2+2t+5}\le\dfrac{32}{25}$$
or$$32(t^4+2t^3+6t^2+2t+5)-25(t+1)^4\ge 0$$
or
$$ (t-3)^2(7t^2+6t+15)\ge 0$$
But this method if we without derivative,we don't known the maximum is $\dfrac{4\sqrt{2}}{5}$.
idea 2
$$f(x)=\dfrac{\sin{x}+1}{\sqrt{(\sin{x}+1)+2(\cos{x}+1)}}$$
Let $u=\sin{x}+1,v=\cos{x}+1$,then $(u-1)^2+(v-1)^2=1$,find the maximum of the
$$\dfrac{u}{\sqrt{u+2v}}$$

Hint:
Assume that $f(x)\leq 4\sqrt2/5$
Now cross multiply, do squares on both sides and ultimately if it's the maximum, you will get an inequality where the equality condition can hold, then write the whole sum backwards!
