<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] better is to show that $\cos(A)+\cos(B)+\cos(C)=1+\frac{r}{R}$ where $r$ is the inradius and $R$ is circumradius of the given triangle. Your theorem follows from this equation
</s>