Optimizing $a+b+c$ subject to $a^2 + b^2 + c^2 = 27$ 
If $a,b,c \gt 0$ and $a^2+b^2+c^2=27$, find the maximum  and minimum values of $a+b+c$.

How to solve this one?
(Here's the source of inspiration for the problem.)
 A: Write $x=(a,b,c),\quad \|x\|^2=27, \quad x\cdot(1,1,1)=(\sqrt{27})(\sqrt{3})\cos\theta$. You can maximize by choosing $\theta=0$, i.e. $a=b=c=3$. You can't minimize the quantity unless you make the bounds on $a,b,c$ non-strict (allow them to take on the value $0$) so that you may choose any of the positive axis vectors for $x$ (as they are farthest away from $(1,1,1)$). This gives a maximum of $9$ and a minimum of $3\sqrt{3}$.
A: You can also use the method of Lagrange multipliers to maximize/minimize $f(a,b,c)=a+b+c$ given $g(a,b,c)=a^2+b^2+c^2=27$. We need the gradient of $f$ to be a multiple of the gradient of $g$, i.e., 
$$1=\lambda 2a,\quad 1=\lambda 2b,\quad 1=\lambda 2c,$$
where $\lambda$ is some real number. Hence:
$$\lambda = \frac{1}{2a}=\frac{1}{2b}=\frac{1}{2c}$$ 
and we must have $a=b=c$. This yields $g(a,a,a)=3a^2=27$, so that $a=b=c=3$ or $a=b=c=-3$. We clearly have a maximum at $a=b=c=3$. Since you assumed $a,b,c>0$, the minimum at $a=b=c=-3$ is not an allowed solution. Thus, the minimum must occur in the boundary of your domain (outside of your domain, so there is an infimum, but no minimum), i.e., when one at least on of $a,b,c$ is zero. However, say $c=0$, then we are trying to minimize $a+b$ given $a^2+b^2=27$. The min occurs when $a=b=-\sqrt{27/2}$ which is again outside our domain $a,b>0$, so the infimum occurs at the boundary when one of $a$ or $b$ is zero. And this leads to an infimum for $f(a,b,c)=a+b+c$ at $(\sqrt{27},0,0)$, $(0,\sqrt{27},0)$ and $(0,0,\sqrt{27})$, and the infimum value is $\sqrt{27}=3\sqrt{3}$. 
A: Since the OP seems to ask for algebraic approaches, here are some proofs based on a symmetrization trick + a bit of algebra for the maximum, and on a similar bit of algebra for the infimum. 
Note $t=27$, $u(a,b,c)=a^2+b^2+c^2$ and $v(a,b,c)=a+b+c$. Let $a$, $b$ and $c$ be positive. Consider $m=\frac13(a+b+c)$ their arithmetic mean. Then 
$$3(u(a,b,c)-u(m,m,m))=(a-b)^2+(b-c)^2+(c-a)^2.
$$
Hence $u(m,m,m)\le u(a,b,c)$ and  $u(m,m,m)<u(a,b,c)$ if $a$, $b$ and $c$ are not all equal. 
Since $u$ is a continuous function, this means that if $a$, $b$ and $c$ are not all equal and if $u(a,b,c)=t$, then there exists $\theta>1$ such that $u(\theta m,\theta m,\theta m)=t$ as well. Since $v(a,b,c)=3m<3\theta m=v(\theta m,\theta m,\theta m)$, $v$ can only be maximum at a point $(x,x,x)$. 
Now, if $u(x,x,x)=t$, $x=\sqrt{t/3}$ and $v(x,x,x)=3\sqrt{t/3}$ hence the maximum is $\sqrt{3t}=9$.
As regards the infimum, 
$$
u(3m,0,0)-u(a,b,c)=2(ab+bc+ca),
$$
hence $u(3m,0,0)=9m^2\ge u(a,b,c)$ and  $u(3m,0,0)>u(a,b,c)$ if $a$, $b$ and $c$ are all positive. Thus, if $u(a,b,c)=t$, $v(a,b,c)=3m>\sqrt{t}$ and the infimum is $\sqrt{t}=3\sqrt3$.
A: Here's a geometric way of looking at it... the points where $x,y,z > 0$ where $x^2 + y^2 + z^2 = 27$ is the "upper right" 1/8 of the outside of the sphere centered at the origin, of radius $3\sqrt{3}$. Call this surface $S$. If you connect the corners $(3\sqrt{3},0,0)$, $(0,3\sqrt{3},0)$, and $(0,0,3\sqrt{3})$ of $S$, you get a triangle in the plane with equation $x + y + z = 3\sqrt{3}$. If you replace $3\sqrt{3}$ by any $r < 3\sqrt{3}$, then the plane $x + y + z = r$ will no longer intersect $S$. So the minimum is $3\sqrt{3}$.
Similarly, the plane $x + y + z = 9$ intersects $S$ at $(3,3,3)$ only, and replacing $9$ by any $r > 9$ will result in a plane   $x + y + z = r$ that doesn't intersect $S$ at all. So the maximum is $9$.
A: The problem could be solved easily and efficiently by the direct use of the Cauchy–Schwarz inequality.
Please check Byron Schmuland answer here.
A: This particular problem can be solved using very elementary techniques of analytic geometry once you realize that $a+b+c=k$ describes, for a fixed value of $k$, a plane. Thus the curves on the sphere $a^2+b^2+c^2=27$ where the function $a+b+c$ is constant are the circles obtaining intersecting it with the said planes.
Thus, finding the maximum value for $a+b+c$ amounts to finding the maximum value of $k$ such that the intersection between the sphere and the plane $a+b+c-k=0$ is non empty. This happens when the plane is tangent and since all these planes are perpendicular to the vector $(1,1,1)$ and since the tangent plane is always perpendicular to the radius at the point of tangency, the point of tangency must be the point $P=(3,3,3)$, so that $k=9$.
On the other hand minimizing $a+b+c$ with the condition $a, b, c\geq0$ is equivalent to finding the smallest value of $k$ such that the same intersection is entirely contained in the first octant. Given the symmetry of the situation, it is quite clear that this "biggest" intersection is the circle joining the points on the sphere which lie on the positive side of the axes, i.e. the points which have two coordinates $0$ and the third equal to $3\sqrt{3}$. So, we get at once $k=3\sqrt{3}$.
