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?

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    $\begingroup$ Probably easier to rephrase without the trig: if $0\leq x,y,z\leq 1$ and $x^2+y^2+z^2=2$ minimize $x+y+z$. $\endgroup$ – Thomas Andrews Aug 25 '13 at 1:43
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    $\begingroup$ @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. $\endgroup$ – chubakueno Aug 25 '13 at 1:48
  • $\begingroup$ 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... $\endgroup$ – Thomas Andrews Aug 25 '13 at 1:54
  • $\begingroup$ @chubakueno Thomas is implying you should try optimization. $\endgroup$ – Don Larynx Aug 25 '13 at 2:04
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    $\begingroup$ Thomas Andrew's equivalent problem seems to succumb quickly to Lagrange multipliers (if you know calculus). $\endgroup$ – Potato Aug 25 '13 at 2:06

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 $f(t)$.

  • $\begingroup$ 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?. $\endgroup$ – chubakueno Aug 25 '13 at 2:20
  • $\begingroup$ 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. $\endgroup$ – zyx Aug 25 '13 at 2:29
  • $\begingroup$ I added and corrected a couple of things. $\endgroup$ – zyx Aug 27 '13 at 6:40

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