<s>[INST] 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!
 [/INST] 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 $0<B,C<180^\circ$
where $\sin\dfrac A2=\dfrac12\implies\dfrac A2=30^\circ$ as $0<\dfrac A2<90^\circ$
</s>