In $ \triangle ABC$ show that $ 1 \lt \cos A + \cos B + \cos C \le \frac 32$

Question

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!

Answer 1

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 $0<\dfrac A2<90^\circ$

Answer 2

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

Answer 3

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:

$$\cos A + \cos B - \cos(A + B)$$ $$=2 \cos \left(\frac{A+B}{2} \right) \cos \left(\frac{A-B}{2} \right) - 2\cos^2 \left(\frac{A+B}{2} \right) + 1 \tag{1}$$ $$=2 \cos \left(\frac{A+B}{2} \right) \left[\cos \left(\frac{A-B}{2} \right) - \cos \left(\frac{A+B}{2} \right)\right]+1$$

$$≤ 2 \cdot \left[\frac{1}{2}\cdot\cos(\frac{A-B}{2}) \right]^2 + 1 \tag{2}$$ $$≤ 2 \cdot \frac{1}{4} + 1 = \boxed{\frac{3}{2}}. \tag{3}$$

$(1)$ 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.

Answer 4

This answer is posted a whole 9 years after OP asked this question so he probably does not need answer at all, but I find this method quite insightful so I am posting this for anyone who stumbles upon this question in the future like I did.

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By Law of Cosines, $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$. Hence, we have: $$\begin{align} &1<\cos A+\cos B+\cos C\leq\frac{3}{2} \\ \iff &1<\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}+\frac{a^2+b^2-c^2}{2ab}\leq\frac{3}{2} \\ \iff &2<\frac{b^2+c^2-a^2}{bc}+\frac{c^2+a^2-b^2}{ca}+\frac{a^2+b^2-c^2}{ab}\leq3 \\ \iff &2abc<a^2b+b^2c+c^2a+ab^2+bc^2+ca^2-a^3-b^3-c^3\leq3abc \\ \end{align}$$

Proof that $a^2b+b^2c+c^2a+ab^2+bc^2+ca^2-a^3-b^3-c^3>2abc$:

By triangle inequality we have $(a+b-c)(b+c-a)(c+a-b)>0$. Expanding this yields our desired result.

Proof that $a^2b+b^2c+c^2a+ab^2+bc^2+ca^2-a^3-b^3-c^3\leq3abc$:

By Schur's Inequality we have $a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)\geq0$. Again, expanding this yields our desired result.

Hence, we have proven that $2abc<a^2b+b^2c+c^2a+ab^2+bc^2+ca^2-a^3-b^3-c^3\leq3abc$, Q.E.D.