If $a+b+c=k$ and $a^2+b^2+c^2 =2k$ what is the maximum value of $k$?

Question

$a,b,c$ 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.

Have you tried a simpler case with, say, two or one parameters? — AlvinL, Jun 29 '20 at 12:52
If $(a,b,c)$ is the coordinates of a point in 3 space, how can you interpret your equations geometrically? — Somos, Jun 29 '20 at 14:21
Inpreting the 2 equations in 3 dimensional space may complicate the problem and I don't know how to do that. — ooiioooiooiiooioooiiooii 123, Jun 29 '20 at 14:28

g.kov · Answer 1 · 2020-06-29T16:35:38.757

3

\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}

edited Jun 29 '20 at 16:35

answered Jun 29 '20 at 15:13

g.kov

13,581

If x=y=z=2 the 2 equations are satisfied and we get a bigger value for k which is 6 – ooiioooiooiiooioooiiooii 123 Jun 29 '20 at 15:43
@ooiioooiooiiooioooiiooii 123: thanks, updated. – g.kov Jun 29 '20 at 16:37

AlvinL · Answer 2 · 2020-06-30T06:20:32.107

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 $k=6$.

score 0 · Answer 3 · answered Jun 29 '20 at 16:36

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 $\color{red}{6}$.

If $a+b+c=k$ and $a^2+b^2+c^2 =2k$ what is the maximum value of $k$?

3 Answers3