Max of cosa + cosb + cosy+ cosa*cosb*cosy

[Math Tutor]

[tex]\alpha+ \beta+ \gamma = 180^\circ[/tex].
Find the maximum of
[tex]\cos\alpha+\cos\beta+\cos\gamma+\cos\alpha\cos\beta\cos\gamma[/tex]

[Guest]

(1) cos(a)+cos(b)+cos(c) = 1+r/R,
(2) R>=2r (Euler relation)
From (1) & (2) => cos(a)+cos(b)+cos(c)<=3/2, and cos(a)+cos(b)+cos(c)=3/2 when cos(a)=cos(b)=cos(c),
then a=b=c=pi/3
It's obviously seen that max(cos(a)*cos(b)*cos(c)) is reached for the same values of a,b,c.
Max(cos(a)+cos(b)+cos(c)+ cos(a)*cos(b)*cos(c))=(cos(pi/3)+cos(pi/3)+cos(pi/3)+cos(pi/3)*cos(pi/3)*cos(pi/3)=1.625

[Guest]

What an awesome solution! You're a genius, man! It took me a while to fully understand it.

[Guest]

For your aesthetic words let me provide you with another solution:

(1) cos(a)^2 + cos(b)^2 + cos(c)^2 + 2*cos(a)*cos(b)*cos(c) = 1 when a + b + c = pi (a nice trigonometric identity)
(2) Jensen's inequality

[Guest]

Man this is not a theorem so it have to be proved:
(1) [tex]cos(a)^2 + cos(b)^2 + cos(c)^2 + 2*cos(a)*cos(b)*cos(c) = 1[/tex] when [tex]a + b + c = \pi[/tex] (a nice trigonometric identity)

[Guest]

You're right my friend that most of people are really in trouble when needed to solve such an easy relation. I didn't
post a proof because i didn't think that the above relation cannot be proved by readers. I can provide with multiple
answers but i prefer to post now one of the uglier ways and wait from you to give the fastest and the most beautiful
solution here. It could be a "Math Problem of the Week", too.

Here you may apply the law of cosines (where a,b,c - the sides of the triangle)

((b^2+c^2-a^2)/(2 b c))^2+((a^2+b^2-c^2)/(2 a b))^2+((a^2+c^2-b^2)/(2 a c))^2+
2×(b^2+c^2-a^2)/(2 b c)×(a^2+b^2-c^2)/(2 a b)×(a^2+c^2-b^2)/(2 a c) = 1
(From here i'm convinced you can do it on your own --- it's worth to do a bit of effort)

[Guest]

Notice one more interesting thing here as regards max(cos(a) + cos(b) + cos(c)) and max(cos(a) + cos(b) + cos(c)+ cos(a)*cos(b)*cos(c)), when a+b+c=pi:
The maxima are caught within the ratios of successive Fibonacci numbers: 1/1, 2/1, 3/2, 5/3, 8/5, 13/8,...
I ask you whose maxima are 5/3, 8/5 by using the same basic function and the same relation a+b+c=pi?

[Guest]

Good solution!

[Guest]

Another solution starts out from the fact that the inequalities below are known:

(1) cos(A) + cos(B) + cos(C) <= 3/2
(2) -1 < cos(A)cos(B)cos(C) <= 1/8 (this can be proved in many many ways --- one of them by using
the first relation and AM - GM inequality)