Question

For a triangle ABC, prove that
$\cos A + \cos B + \cos C \le \dfrac{3}{2}$
In the case of equality, the triangle will be equilateral.

Answer 1

We will begin by taking the LHS of the inequality as some variable $x$. We will convert the sum of the first two terms into a product. We will use half-angle identity for the third term. Then, we will form a quadratic equation and finally, we will prove the inequality.

Formula used:
We will use the following formulas:

1. Sum to product rule: $\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
2. Half-angle identity: $\cos A = 1 - 2{\sin ^2}\dfrac{A}{2}$

Complete step by step solution:
We are required to prove that $\cos A + \cos B + \cos C \le \dfrac{3}{2}$.
Also, we are supposed to prove that if $\cos A + \cos B + \cos C = \dfrac{3}{2}$, then the triangle is an equilateral triangle.
Let us consider the LHS as some variable $x$ i.e., $\cos A + \cos B + \cos C = x$ ……….$(1)$
We will convert the sum of the first two terms into a product. We get
$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
Now, we will convert the last term i.e., $\cos C$ using a half-angle identity. We have
$\cos C = 1 - 2{\sin ^2}\dfrac{C}{2}$
Using these in equation $(1)$, we get
$2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) + 1 - 2{\sin ^2}\dfrac{C}{2} = x$ ………$(2)$
We know that in a triangle, the sum of the angles is $180^\circ$or $\pi$ radians. i.e., $A + B + C = \pi$. From this, we get $A + B = \pi - C$. Substituting this in equation $(2)$, we have
$2\cos \left( {\dfrac{{\pi - C}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) + 1 - 2{\sin ^2}\dfrac{C}{2} = x$ ………$(3)$
We know that $\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta$. Using this in equation $(3)$, we get
$2\sin \dfrac{C}{2}\cos \left( {\dfrac{{A - B}}{2}} \right) + 1 - 2{\sin ^2}\dfrac{C}{2} = x$
Let us write the above equation as a quadratic equation in $\sin \dfrac{C}{2}$. We have
$- 2{\sin ^2}\dfrac{C}{2} + 2\sin \dfrac{C}{2}\cos \left( {\dfrac{{A - B}}{2}} \right) + 1 - x = 0$
Multiplying throughout the above equation by $( - 1)$, we get
$2{\sin ^2}\dfrac{C}{2} - 2\sin \dfrac{C}{2}\cos \left( {\dfrac{{A - B}}{2}} \right) - 1 + x = 0$ ……..$(4)$
Equation $(4)$ is a quadratic equation in $\sin \dfrac{C}{2}$. Since $\sin \dfrac{C}{2}$ is real, the discriminant of the above
equation is greater than or equal to zero.
The discriminant of equation $(4)$ is $\Delta = {b^2} - 4ac$ i.e., $\Delta = {\left( {2\cos \dfrac{{A - B}}{2}} \right)^2} - 4(2)(1 - x)$. Simplifying this discriminant, we get
$\Delta = 4{\cos ^2}\left( {\dfrac{{A - B}}{2}} \right) - 4(2x - 2) \ge 0$
Dividing throughout the inequality by 4, we get
${\cos ^2}\left( {\dfrac{{A - B}}{2}} \right) - (2x - 2) \ge 0$
Transposing the term $2x - 2$ to the RHs of the inequality, we get
${\cos ^2}\left( {\dfrac{{A - B}}{2}} \right) \ge (2x - 2)$ ………$(5)$
We know that $\cos \theta \le 1$. So, ${\cos ^2}\left( {\dfrac{{A - B}}{2}} \right) \le 1$. Using this in inequality $(5)$, we have
$2x - 2 \le 1$
$\Rightarrow x \le \dfrac{3}{2}$ ……….$(6)$
From equation $(1)$and inequality $(6)$, we get
$\cos A + \cos B + \cos C \le \dfrac{3}{2}$
Now, we have to show that if $\cos A + \cos B + \cos C = \dfrac{3}{2}$, then the triangle is equilateral. An equilateral triangle is one in which all sides and all angles are equal.
Consider $\cos A + \cos B + \cos C = \dfrac{3}{2}$
We can write $\dfrac{3}{2}$ as $3 \times \left( {\dfrac{1}{2}} \right)$ which is $\dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2}$.
So, $\cos A + \cos B + \cos C = \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2}$.
Let us take $\cos A = \dfrac{1}{2}$, $\cos B = \dfrac{1}{2}$ and $\cos C = \dfrac{1}{2}$.
Now, to find the angles $A,B$ and $C$, we have $A = {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right)$, $B = {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ and $C = {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
Since $A,B$ and $C$ are acute angles, the only possibility is that ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = 60^\circ$. Hence, $A = B = C$

Therefore, the triangle is an equilateral triangle.

Note:
For the equality part, we can use an alternate method. We will use the law of cosines and the fact that the sum of two sides of a triangle is always greater than the third side. Law of cosines states that “If $A,B$ and $C$ are the angles of a triangle $ABC$, and $a,b,c$ are the sides opposite to the angles $A,B$ and $C$ respectively, then
$\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$, $\cos B = \dfrac{{{c^2} + {a^2} - {b^2}}}{{2ac}}$, and $\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$”
Also, $a + b > c$, $c + b > a$ and $a + c > b$.