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

Hint:
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 \[a + c > b\].