Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This problem is a problem in the last selection phase of the math olympiads in my country.

If $\alpha, \beta,\gamma$ are angles $\in[0,\frac\pi2]$ such that $\sin^2(\alpha)+\sin^2(\beta)+\sin^2(\gamma)=1$.Minimize $\cos(\alpha)+\cos(\beta)+\cos(\gamma)$

I started by $$\cos^2(\alpha)+\cos^2(\beta)+\cos^2(\gamma)=2$$ Then, how can I minimize it? According to Wolfram Alpha, the anwer is $2$ for $(0,1,1)$ and permutations. Just squaring the desired value does not help, what can I do then?

Another additional thought about the problem is that if we take the solutions by Wolfram as true, this hints us that inequalities like $AM\ge GM$ are very unlikely to help.

Major edit

We can show that for every $a \in [0,1]$ $$a^2\le a$$ Then, we use that three times and show we can achieve equality. Is this proof right?

share|improve this question
8  
Probably easier to rephrase without the trig: if $0\leq x,y,z\leq 1$ and $x^2+y^2+z^2=2$ minimize $x+y+z$. –  Thomas Andrews Aug 25 '13 at 1:43
1  
@ThomasAndrews That was the way it was posed(but in spanish), so I wrote it like this to keep most of the essence of the problem. The way you say is a more algebraic and succint (and probably better) to pose the problem. –  chubakueno Aug 25 '13 at 1:48
    
I didn't mean that my way to pose it is better, but that realizing that my problem was equivalent might make it easier to solve... –  Thomas Andrews Aug 25 '13 at 1:54
    
@chubakueno Thomas is implying you should try optimization. –  Don Larynx Aug 25 '13 at 2:04
1  
Thomas Andrew's equivalent problem seems to succumb quickly to Lagrange multipliers (if you know calculus). –  Potato Aug 25 '13 at 2:06

1 Answer 1

up vote 2 down vote accepted

Let $f(t) = \sqrt{1 - t}$. The problem is to minimize $f(a)+f(b)+f(c)$ when $a,b,c \in [0,1]$ and $a+b+c=1$. The graph of $f(t)$ is concave, a piece of the upper half of the parabola $y^2 = 1-x$. As for any concave function, the maximum is attained when $a=b=c$ and the minimum is attained when $a,b,c$ are all [or all except one of them, if all is not possible] equal to $0$ or $1$, which happens at permutations of $(1,0,0)$.

The proof in the edit is correct, but limited to the case where the value of $(a+b+c)$ is consistent with all the values being at ends of the interval. It is the principle that a concave function is $\geq$ the linear function between any two of the points on its graph. In this application, the function is $g(t)=\sqrt{t}$, and the same could have been done directly for $f(t)$.

share|improve this answer
    
Can you please justify your "As for any concave function..." step? Or at least point me to a good source about functions, convexity and concavity?. –  chubakueno Aug 25 '13 at 2:20
    
I don't know a source, but Wikipedia for convexity, convex function, or Jensen inequality should have it. Or a general web search for those terms. It is one of the basic properties. –  zyx Aug 25 '13 at 2:29
    
I added and corrected a couple of things. –  zyx Aug 27 '13 at 6:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.