<s>[INST] 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
$$\dfrac{u}{\sqrt{u+2v}}$$
 [/INST] Change variables so that $\sin(x)$ is $x$ and $\cos(x)$ is $\sqrt{1-x^2}$. The range of $x$ is $[0,1]$, of course. 
Square the function to get rid of the radical, so that now your function is:
$(1+x)^2/(3+x+2\sqrt{1-x^2})$.
Try to prove by direct manipulation that $(1+x)^2/(3+x+2\sqrt{1-x^2})]\leq1$ in $[0,1]$ , i.e. get all but the radical on one side, then get rid of the radical and arrive at a polynomial inequality. You will be able to factor it and show that it is strictly negative.
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