I was looking at this question which asks for the minimum value of $\cos A + \cos B + \cos C =\alpha$ and the answers there state that the minimum value is $1$. This value exists for a degenerate triangle.

But in a similar question which asks for the minimum area ($A$), for a given semi-perimeter ($s$), the possible values of $A$ are

$$\frac {s^2}{3√3} \tag 1$$

in the case when $a=b=c$,


$$\frac {s^2}{4} \tag 2$$

in the case when $s=s-a=s-b=s-c$.

But we reject $(2)$ here as then the triangle wouldn't be an "appropriate" triangle (as my teacher said).

Even though in both $\alpha$'s as well as $(2)$'s case the triangles become degenerate but one of them is acceptable and the other is not. Why so?

Also, if possible please point out the corrections to any conceptual mistake that I may have.

Thank you!

  • $\begingroup$ In $\Delta ABC$, $$\cos A+\cos B+\cos C=1+4\sin\frac A2\sin\frac B2\sin \frac C2$$ $\endgroup$ Jun 28 '20 at 6:25
  • $\begingroup$ The expression (1) gives the maximum area. The origin of the expression (2) should be clarified as the given explanation implies $a=b=c=s=0$. $\endgroup$
    – user
    Jun 28 '20 at 6:50

We'll begin with a proof, possibly not the shortest possible one, of an identity @HarishChandraRajpoot quoted:$$\begin{align}\cos A+\cos B-\cos(A+B)&=1+\sin A\sin B-(1-\cos A)(1-\cos B)\\&=1+4\sin\tfrac{A}{2}\sin\tfrac{B}{2}\cos\tfrac{A+B}{2}\\&=1+4\sin\tfrac{A}{2}\sin\tfrac{B}{2}\sin\tfrac{C}{2}.\end{align}$$This can be made arbitrarily close to $1$, but must exceed it for a nondegenerate triangle, so $1$ is an infimum but not a minimum obtainable in nondegenerate triangles. If $A,\,B$ are both small, to quadratic order$$\cos A+\cos B-\cos(A+B)\approx1-A^2/2+1-B^2/2-1+(A+B)^2/2=1+AB.$$On the other hand, the supremum is an achievable maximum. Since$$\frac{d^2}{dx^2}\ln\sin x=\cot^\prime x=-\csc^2 x<0,$$by Jensen's inequality$$\ln\sin\tfrac{A}{2}+\cdots\ge3\ln\sin\frac{\pi}{6}=-3\ln 2\implies 1+4\sin\tfrac{A}{2}\cdots\ge1+4e^{-3\ln2}=\tfrac32,$$with equality iff the triangle is equilateral. (Note that if we'd tried using Jensen's inequality directly on a sum of cosines, the second derivative $-\cos x$ would have had a different sign for obtuse angles, which is why we had to work in terms of half-angles.)


Does not exist.

For $a=b=1$ and $c\rightarrow2^-$ we see that $$\cos\alpha+\cos\beta+\cos\gamma\rightarrow1.$$

We'll prove that $$\cos\alpha+\cos\beta+\cos\gamma>1.$$ Indeed, we need to prove that: $$\sum_{cyc}\frac{b^2+c^2-a^2}{2bc}>1$$ or $$\sum_{cyc}(a^2b+a^2c-a^3)>2abc$$ or $$(a+b-c)(a+c-b)(b+c-a)>0,$$ which is obvious.

The value $1$ does not occur, which says that the minimal value does not exist.

Now, we see that $\sum\limits_{cyc}\cos\alpha\rightarrow1^+$ for any $a+b\rightarrow c^+$.

Your case with $p-a=p-b=p-c\rightarrow p$ is impossible because $a$, $b$ and $c$ are positives.

For $a=b=c$ we'll get a maximal value of the sum.

  • $\begingroup$ "Does not exist." I believe it's regarding the title? So do you mean that the answer on the referred post (in my question) is wrong? If so thanks. (BtW I didn't downvote) $\endgroup$
    – user743391
    Jun 28 '20 at 8:21
  • $\begingroup$ @Johan Liebert Yes, I answered on the question in the title. $\endgroup$ Jun 28 '20 at 8:58

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