Proving Hard Inequalities for Acute Angles and Trigonometric Functions

[MathematicalPhysicist]

i need to prove the following:
1)let a,b,c be acute angles, if tg(a)tg(b)tg(c)=1 then sin(a)sin(b)sin(c)<=1/2sqrt2
2) prove that for every x,y cos(x^2)+cos(y^2)-cos(xy)<3
for the second question i tried to use the fact that (x^2+y^2)/2>=xy and the fact that on some intervals the function cos is decreasing, actually what i need to prove is that -cos(xy)<1, cause cosx^2+cosy^2<=2, so i also tried to show that cos(xy) cannot be equal -1, but didnt get much with that.

for the first i used the cosine law and sine law, but also didnt get far, i need to show that (sin^2a(sin^2b+sin^2c)+sin^2b(sin^2a+sin^2c)+sin^2c(sin^2a+sin^2b)<=sin(a)sin(b)sin(c), but i didnt succeded in it.

p.s
if someone wonders why I am posting questions of these type is because I am reviewing a little bit inequalities.

 

[matt grime]

tg(a) is what?

 

[MathematicalPhysicist]

a shortcut for tangans of a, or if you will sina/cosa. (-:

 

[matt grime]

It's tan, a well known abbreviation already. Is this some new convention with which I'm completely unfamiliar?

 

[MathematicalPhysicist]

as far as I am aware you can use both tan and tg.
anyway, do you have any hints about my questions?

 

[Office_Shredder]

For the second one, the maximum value cos can attain is 1, and the minimum is -1. So you simply have to show that cos(x²)=cos(y²)=1 and cos(xy)=-1 are not simultaneously possible, because that's the only way to make that equal three (you can't break three with the sum of three cos terms). So for cos to equal 1,x^2=2\pi n and same for y² Try to analyze what xy can be given that (and noting xy must be a multiple of (2n+1) \pi to make your inequality untrue)

 

[arildno]

Have you tried using a Lagrange multiplier approach on the first one?

 

[matt grime]

I thought of that too, but it seems a bit too complicated, if you ask me. Here are two proofs of the first one without any analysis/calculus.

The condition essentially states that sin(a)sin(b)sin(c)=cos(a)cos(b)cos(c). Or that the function is symmetric about (pi/4,pi/4,pi/4) in each variable. Thus that is either going to be the maximum or minimum, a check tells us it is not a minimum.

If that's too hand wavy, how about this one? Fix a. Clearly for any a the maximum we can attain is when b=c. This applies for any a, in particular the a where the maximum occurs. Thus the maximum occurs at (a,b,b). But this is completely symmetric in the arguments, so a=b=c is the maximum. The value of the maximum is then gotten from the condition which reduces to tan(a)^3=1, or a=pi/4.

I can't be bothered to make those rigorous - which could be done with an argument involving lagrange multipliers, for example.

 

[Office_Shredder]

matt, I think your first argument is less hand wavy than the second, which actually assumes off the bat that there DOES exist a maximum.

 

[matt grime]

But there is obviously a maximum by inspection.

 

[arildno]

As it happens, using Lagrange multipliers turns out to be very simple.
It is, however, at least as important to understand how the symmetry arguments used by matt are powerful, and ought to be convincing on their own.

Here's how we might set up the problem using L.M's:
Define:
f(a,b,c)=\sin(a)\sin(b)\sin(c), g(a,b,c)=f(a,b,c)-\cos(a)\cos(b)\cos(c)
Thus, we are to solve the problem:
\nabla{f}=\lambda\nabla{g},g=0
Considering the first equation, we have:
\cos(a)\sin(b)\sin(c)=\lambda(\cos(a)\sin(b)\sin(c)+\sin(a)\cos(b)\cos(c))
Rearranging a bit, we may write this equation as:
\tan(a)=\frac{1-\lambda}{\lambda}\tan(b)\tan(c)
Similar expressions hold, of course, for tan(b) and tan(c), and multiplying these three equations and utilizing g=0, yields:
(\frac{1-\lambda}{\lambda})^{3}=1\to\lambda=\frac{1}{2}
OP can finish the exercise on his own.

 

[MathematicalPhysicist]

the problem is i can't use lagrange multipliers.

here's my appraoch:
we have three acute angles, we can use the cosine law and sine law,
after I am using both of these theorems, i get to this inequality:
sin(a)sin(b)sin(c)<=1/8[sin^2a/sin^2c+sin^2a/sin^2b+sin^2b/sin^2a+sin^2b/sin^2c+sin^2c/sin^2b+sin^2c/sin^2a]
p.s
matt, i feel that both your arguments are hand wavy.
i understand that sin(a)sin(b)sin(c)=cos(a)cos(b)cos(c) but how do you show that the maximum is achieved when a=b=c=pi/4?

 

[MathematicalPhysicist]

office shredder, if cos(xy)=-1 cos(x^2)=cos(y^2)=1 then we have
x^2=2npi y^2=2kpi so xy=2(pi)sqrt(nk) but according to the first equation xy=pi+2cpi so we have 2(pi)sqrt(nk)=pi+2cpi
4nk=1+2c which is a contradiction cause the rhs is an odd number, and the lhs is even.
ok, thanks.