trigonometry - In $triangle ABC$ show that $1 lt cos A + cos B + cos C le frac 32$

Here is what I did, tell me whether I did correct or not:
$$\begin{align*} y &= \cos A + \cos B + \cos C\\ y &= \cos A + 2\cos\left(\frac{B+C}2\right)\cos\left(\frac {B-C}2\right)\\ y &= \cos A + 2\sin\left(\frac A2\right)\cos\left(\frac {BC}2\right) && \text{since $A+B+C = \pi$} \end{align*}$$

Now for maximum value of $y$ if we put $\cos\left(\frac {B-C}2\right) = 1$ then
$$\begin{align*} y &\le \cos A + 2\sin\left(\frac A2\right)\\ y &\le 1-2\sin^2\left(\frac A2\right) + 2\sin\left(\frac A2\right) \end{align*}$$

By completing the square we get

$$y \le \frac 32 - 2\left(\sin\frac A2 - \frac 12\right)^2$$

$y_{\max} = \frac 32$ at $\sin\frac A2 = \frac 12$ and $y_{\min} > 1$ at $\sin \frac A2>0$ because it is a ratio of two sides of a triangle.

Is this solution correct? If there is a better solution then please post it here. Help!

Answer

We can prove $$\cos A+\cos B+\cos C=1+4\sin\frac A2\sin\frac B2\sin\frac C2$$

Now, $0<\dfrac A2<90^\circ\implies\sin\dfrac A2>0$

For the other part,

$$y=1-2\sin^2\frac A2+2\sin\frac A2\cos\frac{B-C}2$$

$$\iff2\sin^2\frac A2-\sin\frac A2\cdot2\cos\frac{B-C}2+y-1=0$$ which is a Quadratic equation in $\sin\dfrac A2$ which is real

So, the discriminant must be $\ge0$

i.e., $\left(2\cos\dfrac{B-C}2\right)^2-4\cdot2(y-1)\ge0$

$\iff 4y\le4+2\cos^2\dfrac{B-C}2=4+1+\cos(B-C)\le4+1+1$

The equality occurs iff $\cos(B-C)=1\iff B=C$ as $0

where $\sin\dfrac A2=\dfrac12\implies\dfrac A2=30^\circ$ as $0<\dfrac A2<90^\circ$