If $a+b+c=k$ and $a^2+b^2+c^2 =2k$ what is the maximum value of $k$? $a,b,c$ are real numbers and they satisfy the following equations.
$a+b+c=k$
$a^2+b^2+c^2=2k$
Find the maximum value of $k$.
I tried substituting for k in the second equation from the first and got
$a^2+b^2+c^2=2(a+b+c)$
Rearranging the terms I got
$a^2-2a+b^2-2b+c^2-2c=0$
Adding 3 to both sides we get
$a^2-2a+1+b^2-2b+1+c^2-2c+1=3$
This can be simplified to the following
$(a-1)^2+(b-1)^2+(c-1)^2=3$
Therefore,
$0\leq(a-1)^2,(b-1)^2,(c-1)^2\leq3$
From here we can deduce the range of values that a,b,c can take as
$1-\sqrt{3}\leq a,b,c\leq1+\sqrt{3}$
I don't know know if this helps to answer the question.
 A: 
\begin{align} x+y+z&=k \tag{1}\label{1},\\ x^2+y^2+z^2&=2k\tag{2}\label{2}. \end{align}

Expressing $y,z$ in terms of $x,k$ gives
\begin{align}
y,z&=
\tfrac12\,(k-x\pm\sqrt{-k^2+2 x k-3x^2+4k})
,
\end{align}
so we must have
\begin{align}
-k^2+2 x k-3x^2+4k\ge0
,
\end{align}
which leads to expression of $k$ in terms of $x$
\begin{align}
k(x)&=x+2+\sqrt{4+4x-2x^2}
,\\
k'(x)&=
\frac{\sqrt{4+4x-2x^2}+2-2x}{\sqrt{4+4x-2x^2}}
,
\end{align}
$k'(x)=0$ at $x=2$.
\begin{align}
k_{\max}&=k(2)=6
.
\end{align}
A: Your ideas are good. You obtained $(a-1)^2 + (b-1)^2 + (c-1)^2 = 3$. That means there is a sphere around the point $(1,1,1)$ with radius $\sqrt{3}$ and the point $(a,b,c)$ is situated on that sphere.
The condition $a^2+b^2+c^2 = 2k$ tells us that the point $(a,b,c)$ lives on a sphere around the origin at distance $\sqrt{2k}$.
The point exists on both spheres simultaneously, hence ...? (what is the largest distance from the origin these spheres can intersect?)

2D analogue. The point $C$ in the picture corresponds to $(a,b)$. Then $2\sqrt{2} = \sqrt{2k}$ i.e $k=4$.

In the 3D case, we would have $2\sqrt{3} = \sqrt{2k}$, thus $k=6$.
A: Since $c=k-(a+b)$ both inequalities are fulfilled iff the set of points $(a;b)$ such that
$$ a^2 + b^2 + (k-(a+b))^2 = 2k $$
is non-empty. This equation can be written as
$$ 2a^2+2ab+2b^2-2ka-2kb+(k^2-2k) = 0 $$
or, by letting $a=A+\frac{k}{3},b=B+\frac{k}{3}$, as
$$ 2A^2+2AB+2B^2 = 2k-\frac{k^2}{3}.$$
The matrix $\left(\begin{smallmatrix}2 & 1 \\ 1 & 2\end{smallmatrix}\right)$ is positive definite, hence this is the equation of an ellipse provided that $2k-\frac{k^2}{3}>0$.
It follows that the maximum value of $k$ is $\color{red}{6}$.
