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!
 A: 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 $0<\dfrac A2<90^\circ$
A: 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
A: Since $A,B,C$ are the angles of a triangle, $C = \pi - (A+B)$. WLOG let $A,B$ be acute angles, as there cannot be two obtuse angles in a triangle, which leaves three acute angles or one obtuse, two acute only.  Then $\cos A +\cos B + \cos C$ equals:
$$\cos A + \cos B - \cos(A + B)$$
$$=2 \cos \left(\frac{A+B}{2} \right) \cos \left(\frac{A-B}{2} \right) - 2\cos^2 \left(\frac{A+B}{2} \right) + 1 \tag{1}$$
$$=2 \cos \left(\frac{A+B}{2} \right) \left(\cos \left(\frac{A-B}{2} \right)  - \cos \left(\frac{A+B}{2} \right)\right)+1$$
$$≤ 2 \cdot \left( \frac{\cos((A-B)/2)}{2} \right)^2 + 1 \tag{2}$$
$$≤ 2 \cdot \frac{1}{4} + 1 = \boxed{\frac{3}{2}}. \tag{3}$$
where $(1)$ uses sum-to-product and the double-angle identities, $(2)$ is an application of AM-GM and $(3)$ relies on $\cos u ≤ 1, u \in \mathbb R$.
AM-GM is justified in $(2)$ as $0 < \frac{A+B}{2} < \frac{\pi}{4}$ so the first term is always positive. As for the other term, this is equal to $2 \sin A/2 \sin B/2$ by product-to-sum (or just expand the brackets), which is always nonnegative as both terms are nonnegative.
From this, the minimum value is actually $0 + 1 = 1$, which occurs when $A = B = \pi, C = 0$ which is a degenerate triangle. The maximum value occurs when $\cos \left(\frac{A-B}{2} \right) = 1 \implies \frac{A-B}{2} = 0$. Then setting both terms equal, $2 \cos \left(\frac{A+B}{2} \right) = 1 \implies \cos A = \frac{\pi}{3}$, so when $ABC$ is equilateral.
