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
$$\dfrac{u}{\sqrt{u+2v}}$$
 A: 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.
A: We need to minimize $$\dfrac{3+2\cos x+\sin x}{(1+\sin x)^2}$$
Now WLOG let $x=\dfrac\pi2-2y$ to get $$\dfrac{3+2\sin2y+\cos2y}{(1+\cos2y)^2}$$
Using Weierstrass substitution, writing $\tan y=t$
we get $$2f(t)=(t^2+1)(t^2+2t+2)$$
Now use Second derivative test,  to find the minimum value of $f(t)$ occurs at $-\dfrac12$
i.e., $$f(t)\le\dfrac{25}{32}$$ 
A: 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!
