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?

• 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
• @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
• Thomas Andrew's equivalent problem seems to succumb quickly to Lagrange multipliers (if you know calculus). – 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 $f(t)$.