Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.)

share|cite|improve this question
Did you try drawing a picture? Also, are you sure you want $a,b,c \gt 0$ and not $a,b,c \geq 0$? – t.b. Aug 23 '11 at 10:25
To check myself before I go on, the answers are 9 and the square root of 27, yes? – mixedmath Aug 23 '11 at 10:25
@FoolForMath The thing about the minimum is that if a,b,c are allowed to be zero, then that's grand. But if not, my minimum is incorrect and it's not well-defined (there's a greatest lower bound, but no minimum). – mixedmath Aug 23 '11 at 10:36
The thing is: $a^2 + b^2 + c^2 = 27$ is the sphere with radius $3\sqrt{3}$ and $a,b,c \geq 0$ cuts out the part in the positive octant. The maximum value will be achieved when $a=b=c$ and the minimum when two of them are zero. Your constraint $a,b,c \gt 0$ will lead to an infimum, not a minimum. – t.b. Aug 23 '11 at 10:39
@Theo Buehler:Thanks for all your inputs :-) – Quixotic Aug 23 '11 at 20:39

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 $3\sqrt{3}$.

share|cite|improve this answer
You lost me at this $x=(a,b,c),\quad \|x\|^2=27, \quad x\cdot(1,1,1)=(\sqrt{27})(\sqrt{3})\cos\theta$ ... I don't understand sorry. – Quixotic Aug 23 '11 at 10:43
@FoolForMath: Do you know what vectors or the dot product are? – anon Aug 23 '11 at 10:46
I have studied them before but didn't really used any of it in recent time.Btw this question is CAT standard which do not have have vectors in the syllabus so I guess there must be another trivial approach. – Quixotic Aug 23 '11 at 10:50
... I do understood this approach now,Thanks for those links:-) – Quixotic Aug 23 '11 at 10:57
@FoolForMath: The brute force analytic approach would be to use Lagrange multipliers (which will be useful later in studying economics) the linked example deals with the case of two variables (without positivity constraints) and it is already more painful than what anon posted, so you can imagine how the three-dimensional case will look. – t.b. Aug 23 '11 at 11:05

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 $\sqrt{27}=3\sqrt{3}$.

share|cite|improve this answer
Nicely done. This is the way I like to prove these kinds of problems. It definitely shows, without any doubt, what the maximum and minimum are. – robjohn Aug 23 '11 at 16:30

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 $\sqrt{t}=3\sqrt3$.

share|cite|improve this answer

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 $9$.

share|cite|improve this answer
up vote 0 down vote accepted

The problem could be solved easily and efficiently by the direct use of the Cauchy–Schwarz inequality.

Please check Byron Schmuland answer here.

share|cite|improve this answer
Yes Cauchy-Schwarz yields the maximum (and the argument is used in my answer, by the way, hence if you read this answer it will get you an algebraic proof of Cauchy-Schwarz inequality for 3D vectors, gratis pro bono). The infimum seems to require a different argument than Cauchy-Schwarz. – Did Aug 23 '11 at 20:19

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 $k=3\sqrt{3}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.