Minimum value of $\displaystyle f(x) = \frac{2+\sin x}{2+\cos x}$.
My try: let $$\displaystyle y = \frac{2+\sin x}{2+\cos x}\Rightarrow 2y+y\cdot \cos x = 2+\sin x$$
So $$y\cdot \cos x-\sin x= 2-2y\;,$$ Now using Cauchy-Schwarz Inequality::
$$\displaystyle \left(y^2+(-1)^2\right)\cdot (\cos^2 x+\sin^2 x)\geq (y\cos x-\sin x)^2\Rightarrow (y^2+1)\geq (2-2y)^2$$
So after simplifying $$\displaystyle y^2+1\geq 4+4y^2-8y\Rightarrow 3y^2-8y+3\leq 0$$
So $$\displaystyle 3y^2-8y+3\leq 0 \Rightarrow \frac{1}{2}(4-\sqrt{7})\leq y\leq \frac{1}{3}\left(4+\sqrt{7}\right)$$
But answer given is $\displaystyle = \frac{2}{3}$
So please explain where I am wrong, thank you.