Prove that $\cos A + \cos B +\cos C \le \frac{3}{2} $ using Jensen's Inequality where $A,B,C$ are angles of a triangle

Question

I came across this question in an exercise related to Jensen's Inequality, but I know that the function $ f(x) = \cos x$ changes its concavity after $\frac{\pi}{2}$. So I was not able to prove this result for an obtuse-angled triangle. So I want to know how this question can be solved using Jensen's Inequality. I was able to solve this question using a different approach here:-
Let, $\cos(A) + \cos(B) + \cos(C) = k.$
On factorizing $ \cos A + \cos B $ and applying 2 times angle formula for $\cos C$, we get
$$ 2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right) + 2\cos^2 \left(\frac{C}{2}\right)-1 = k.$$
As $A+B = \pi - C$, we get:-

$$2\sin \left(\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right) + 2\cos^2 \left(\frac{C}{2}\right)-1.$$

On changing $2\cos^2 \left(\frac{C}{2}\right)$ into $\sin$ and simplifying, we get a quadratic in $\sin \left(\frac{C}{2}\right)$, Letting $\sin \left(\frac{C}{2}\right)$ as $t$, we get:-

$$t^2 - \left(2\cos \left(\frac{A-B}{2}\right)\right)t + (k-1) = 0.$$
For this to have real roots, its discriminant must be greater than zero, so:
$$ 2k - 2 \le \cos^2 \left(\frac{A-B}{2}\right).$$
Since maximum value of $\cos^2 \left(\frac{A-B}{2}\right)$ is $1$, we get a maximum value of $k = \frac{3}{2}$.

Answer 1

I'm assuming you did the case when all angles are acute, as that's standard Jensens and you mentioned checking concavity. To continue in the spirit of your solution for the obtuse case:

WLOG let $ C \geq \frac{\pi}{2}$, so $ \cos C \leq 0$.

By Jensen's, since $A, B \in [0, \frac{\pi}{2}]$,

$$ \cos A + \cos B \leq 2 \cos \frac{ A+B}{2}.$$

Let $ x = \frac{A+B}{2}$, then we want to maximize

$$\cos A + \cos B + \cos C \leq 2 \cos x - \cos 2x = 1 + 2 \cos x - 2\cos^2 x .$$

Let $ y = \cos x$, then we want to maximize

$$ 1 + 2y - 2y^2 = -2(y - \frac{1}{2})^2 + \frac{3}{2}. $$

Hence, naively $\frac{3}{2}$ is still an upper bound.

Note: Because $ \frac{A+B}{2} \in [0, \frac{\pi}{2}]$, the domain of $y$ is restricted to $ [ \frac{ \sqrt{2}}{2}, 1 ]$, which gives a maximum of $\sqrt{2} $ when $ y = \frac{ \sqrt{2}}{2}$ in this obtuse case.

Answer 2

Hint:

Use the identity $$\cos(A)+\cos(B)+\cos(C)\\=1+4\sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right)$$ The angles are now less than $\frac{\pi}{2}$.

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Let $f(x)=\log(\sin(x))$. $f$ is a concave function for $x\in (0,\frac{\pi}{2}]$, as $$f''(x)=-1-\frac{1}{\tan(x)^2}$$

We may now use Jensen to arrive at $$\frac{1}{3}\sum_{\sigma} \log\left (\sin\left (\frac{A}{2}\right)\right)\leq\sum_\sigma\log\left (\sin\left (\frac{1}{3}\cdot \frac{A+B+C}{2}\right)\right)=\log\left (\frac{1}{2}\right)$$ We then get $$\sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right)\leq\frac{1}{8}$$ and so the claim follows.