Find $x,y,z>0$ such that $x+y+z=1$ and $x^2+y^2+z^2$ is minimal How can I find $3$ positive numbers that have a sum of $1$ and the sum of their squares is minimum?
So far I have:
$$x+y+z=1 \qquad \implies \qquad z=1-(x+y)$$
So,
$$f(x,y)=xyz=xy(1-x-y)$$
But I'm stuck from here.  Hints?
 A: $x+y+z = 1$ is a plane intersecting the positive quadrant in an equilateral triangle through $(0,0,1), (0,1,0), (1,0,0)$. Minimising the sum of squares of the numbers corresponds to finding the point in the triangle closest to the origin. By symmetry, $x=y=z=\dfrac{1}{3}$.
A: $$f(x,y)=x^2+y^2+(1-x-y)^2$$
$$\frac{\partial f}{\partial x}=2x+2(1-x-y)(-1)=0\tag 1$$
$$\frac{\partial f}{\partial y}=2y+2(1-x-y)(-1)=0\tag 2$$
by subtracting 
$$x=y$$ 
substitute in 1 to get
$$x=\frac{1}{3}$$ 
so
$$y=\frac{1}{3}$$
and then $$z=1-x-y=\frac{1}{3}$$ 
to check the critical point, we can reduce the original function, which has three variables, to one varible as follow
$$f(x)=x^2+x^2+(1-x-x)^2=2x^2+(1-2x)^2$$ 
it is very easy to find the critical point if there is one variable
A: $x \mapsto x^2$ is a convex function. By Jensen's inequality,
$$x^2+y^2+z^2 = 3\left(\frac{x^2+y^2+z^2}{3}\right) \ge 3\left(\frac{x+y+z}{3}\right)^2 = 3\left(\frac{1}{3}\right)^2 = \frac13$$
Since the equality $x^2+y^2+z^2 = \frac13$ is achieved at $x = y = z = \frac13$, this is the solution you seek. 
If you want a more elementary approach, you can use the fact
$$\begin{align}
x^2 + y^2 + z^2 
=& \left(x-\frac13+\frac13\right)^2 + \left(y-\frac13+\frac13\right)^2 + \left(z -\frac13 + \frac13\right)^2\\
=& \left(x-\frac13\right)^2 + \left(y-\frac13\right)^2 + \left(z -\frac13\right)^2\\
&+ \frac23\left[\left(x-\frac13\right) + \left(y-\frac13\right) + \left(z -\frac13\right)\right] + 3\left(\frac13\right)^2\\
=& \left(x-\frac13\right)^2 + \left(y-\frac13\right)^2 + \left(z -\frac13\right)^2 + \frac13
\end{align}
$$
to arrive at same conclusion.
A: You want to minimize:
$$min ~ x^2 + y^2 +z^2$$
With restrictions:
$$x,y,z \ge 0$$
$$ x + y +z=1$$
Now to solve this problem you can apply the method of Lagrange multipliers and minimize:
$$x^2+y^2+z^2+ \lambda (x+y+z-1))$$
A: By Cauchy-Schwarz, for every $x,y,z\in\Bbb R$ we have
$$x+y+z=\langle(1,1,1),(x,y,z)\rangle \leq \left\|(1,1,1)\right\|_2\cdot \|(x,y,z)\|_2=\sqrt{3}\cdot \sqrt{x^2+y^2+z^2}$$
Now, if $x+y+z=1$, then squaring both sides and dividing by $3$ gives
$$ \frac{1}{3}\leq x^2+y^2+z^2 $$
This lower bound is then attained for $|x|=|y|=|z|=1/3$. 
A: By elementary geometry. You are looking for a point on the plane given by $x+y+z=1$ that is closest to the origin, which means that the vector $(x,y,z)$ from the origin to that point is a normal vector to the plane. This happens if $x=y=z$, which gives you three independent linear equations in $x,y,z$ to solve.
A: Lagrange multipliers? So $f(x,y,z,\lambda) = x^2 + y^2 + z^2 - \lambda(x+y+z-1)$.
So critical points obey: $2x - \lambda = 0, 2y - \lambda = 0, 2z - \lambda = 0, x+y+z-1 = 0$. So $x = y = z = \frac{1}{3}$, which gives sum of squares $\frac{1}{3}$. 
A: Restating: 
Find x,y,z such that the squared distance from the origin $d^2  := x^2 + y^2 + z^2$ to the plane $g(x,y,z) = x+ y+ z - 1 = 0$ is minimal.
The gradient of  g(x,y,z) is a vector normal to the surface at ( x,y,z).
A normal to the plane:
$\nabla g= (\frac{\partial g}{\partial x},\frac{\partial g}{\partial y}, \frac{\partial g}{\partial z}) = (1,1,1)$.
Consider a vector $ \vec{r}  = a (1,1,1) $ from (0,0) to the plane.
The vector components satisfy   $ x + y + z - 1 = 0,  $
$a + a + a - 1 = 0 , $
$a = 1/3.$
$\vec{r}  =  a (1,1,1) = (1/3,1/3,1/3)$, i.e.  
$ x = y = z = 1/3.$
The min value:
$d^2 = x^2 + y^2 + z^2 = 1/3 $.
A: Another way to see that the optimum is attained at a constant vector is via majorization and the Karamata inequality. 
Credit for pointing this out goes to @zeb in this answer to a question of mine.
A: By C-S $$3(x^2+y^2+z^2)=(1+1+1)(x^2+y^2+z^2)\geq(x+y+z)^2=1.$$
Thus, $x^2+y^2+z^2\geq\frac{1}{3}$.
The equality occurs for $x=y=z=\frac{1}{3}$, which is also the answer.
