TITLE: Another olympiad problem
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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 $f(t)$.