Using Lagrange Multiplier to find global maximum of a bounded function.

Question

Consider this IMO 1984 problem.

Prove that $0≤++−2≤\frac {7}{27}$, where $$,$$ and $$ are non-negative real numbers for which $++=1$.

I have recently learned about Lagrange Multiplier and I intend to use this to solve the above problem.

From what I understand Lagrange Multiplier only gives local maximums/minimums of the bounded function. Secondly, before using it, I must make sure that the function has a maximum/minimum . Third, there can exist points of global maximum/minimum other than the ones found using Lagrange Multiplier.

Provided the above 'limitations', how can I apply Lagrange Multiplier to find the global maximum of the above problem?

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My approach:

I have used Lagrange Multiplier to get local extrema at the points $(\frac 13,\frac 13,\frac 13)$,$(\frac 12,\frac 12,0)$ , $(\frac 12,0,\frac 12)$,$(0,\frac 12,\frac 12)$.

What I am unable to determine is that:

1. Are these the only points of extremum? Can other points exist where there may be global maximum (such as on boundaries of the function)?

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Note that I have finished what is called as "Calculus 1" and "Calculus 2" is some universities but I am yet to formally start "Calculus 3".

Answer 1

You have identified the critical points correctly. Before applying Lagrange Multipliers method, we check whether a maximum and minimum is guaranteed. For that, we apply Extreme Value Theorem (wiki) which states that if a function $f$ in $\mathbb{R}$ is continuous on a closed interval, then it must have a maximum and a minimum in the interval.

As we know, $x+y+z = 1$ is equation of a plane and with condition $x, y, z \geq 0$, it is part of the plane that is in first quadrant bounded by coordinate planes. So we clearly have a closed interval and by Extreme Value Theorem, there must exist a maximum and minimum.

Next thing to remember is that to find maximum and minimum in a closed interval, we must examine both the critical points and the boundary points. While applying Lagrange Multiplier method will give you stationary points of $f$ in the interior of the domain, it may not identify points if the extremum occurs on the boundary. We must examine that separately.

Here $f(x, y, z) = xy + yz + zx - 2 xyz, \ g(x, y, z) = x+y+z-1 = 0$

Applying Lagrange Multiplier method, $f(x, y, z) = \lambda g(x, y, z)$. Taking derivative with respect to $x, y, z, \lambda$,

$x + y - 2 xy = \lambda$
$y + z - 2 yz = \lambda$
$z + x - 2 zx = \lambda$
$x + y +z - 1 = 0$

Solving we get four critical points as you have mentioned in your question.

But $x + y + z = 1$ in first quadrant has $3$ boundary points which is at intersection with three coordinate axes. For example, at intersection with x-axis, $x = 1, y = z = 0$.

So we must examine boundary points $(1, 0, 0), (0, 1, 0), (0, 0, 1)$ in addition to critical points for maximum and minimum. Now due to symmetry in $f(x, y, z)$, we know there are only three points to check for maximum and minimum values and those are $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}), (\frac{1}{2}, \frac{1}{2}, 0)$ and $(1, 0, 0)$.

Answer 2

Consider first for $x,y,z\neq 0$ \begin{align} \begin{cases} f(x,y,z)~=~\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{C}{x+y+z},\\ g(x,y,z)~=~x+y+z \end{cases} \end{align} where $C=\frac{7}{27}$ and the constraint $g(x,y,z)=1$ (so we can divide by $x+y+z$). We have: \begin{align} \begin{cases} \mathrm{d}f(x,y,z)~=~\left(-\frac{1}{x}^2+\frac{C}{(x+y+z)^2}\right) \mathrm{d}x+\left(-\frac{1}{y^2}+\frac{C}{(x+y+z)^2}\right) \mathrm{d}y+\left(-\frac{1}{z^2}+\frac{C}{(x+y+z)^2}\right) \mathrm{d}z,\\ \mathrm{d}g(x,y,z)~=~\mathrm{d}x+\mathrm{d}y+\mathrm{d}z. \end{cases} \end{align} By Lagrange multipliers' theorem, extrema are those $(x,y,z)$ which cancel all the three $2\times2$ sub-determinants of \begin{align} \begin{pmatrix} -\frac{1}{x}^2+\frac{C}{(x+y+z)^2}&-\frac{1}{y}^2+\frac{C}{(x+y+z)^2}&-\frac{1}{z}^2+\frac{C}{(x+y+z)^2}\\ 1&1&1 \end{pmatrix}. \end{align} It is straightforward to see that this happens if and only if \begin{align} \frac{1}{x^2}=\frac{1}{y^2}=\frac{1}{z^2} \end{align} that is $x=\pm y$, $y=\pm z$ and $z\pm x$. The constraint being $g(x,y,z)=1$ and the inequality in the OP requiring $0\leq yz+zx+xy-2xyz$, you find $x=-1,y=z=1$ and permutations (and the inequality in the OP does not hold), or, for positive $x,y,z$, you get $x=y=z=\frac{1}{3}$.

Now consider the three boundaries problems $x=0$ or $y=0$ or $z=0$. Reproduce the operation to obtain all the critical points then the desired inequality; I let you fill the details.

Answer 3

Both the function in question $ \ f(x,y,z) = ++−2 \ \ $ and the constraint plane $ \ ++=1 \ $ have an "exchange symmetry" in that any two variables can be swapped in their expressions without altering them. Under this symmetry, the extremal points will generally have the forms $ \ (x,x,x) \ $ or $ \ (x,x,z) \ , \ (x,y,x) \ $ and $ \ (x,y,y) \ \ . $ This is a manifestation of the shape that the level-surfaces of the function take on; the graph at right above shows the portion of $ \ f(x,y,z) = 0 \ $ in the vicinity of the origin. In the graph at left above is a wider view of the same level-surface, along with the constraint plane $ \ x + y + z = 1 \ \ ; $ there is a part of this surface that lies entirely in the first octant and "sit well above" that plane.

The Lagrange equations, shown in MathLover's answer, can be equated in pairs to produce

$$ \lambda \ \ = \ \ y + z - 2yz \ \ = \ \ x + z - 2xz \ \ \Rightarrow \ \ 2z \ · \ (x - y) \ \ = \ \ x - y \ \ \Rightarrow \ \ (2z - 1) · (x - y) \ \ = \ \ 0 \ \ , $$

and similarly, $ \ (2y - 1) · (x - z) \ \ = \ \ 0 \ \ $ and $ \ (2x - 1) · (y - z) \ \ = \ \ 0 \ \ . $

The only internally consistent groupings of these conditions are to have $ \ x = y = z \ \ ; $ or to have one pair of coordinate variables equal to one another and to $ \ \frac12 \ \ ; $ or one pair equal and the third variable equal to $ \ \frac12 \ \ . $ This tells us that we have points where the gradients for the level-surfaces equal to the normal $ \ \langle 1 \ , \ 1 \ , \ 1 \rangle \ $ at

$ \mathbf{ x = y = z \ : } \quad \left( \frac13 \ , \ \frac13 \ , \ \frac13 \right) \ \ \rightarrow \ \ f\left( \frac13 \ , \ \frac13 \ , \ \frac13 \right) \ = \ 3 · \left(\frac13 \right)^2 \ - \ 2 · \left(\frac13 \right)^2 \ \ = \ \ \frac{3·3 \ - \ 2}{27} \ \ = \ \ \frac{7}{27} \ \ \approx \ \ 0.2593 \ ; $

pair of coordinates $ \ = \ \frac14 \ , $ third $ \mathbf{\ = \ \frac12 \ :} \quad \left( \frac14 \ , \ \frac14 \ , \ \frac12 \right) \ \ , \ \ \left( \frac14 \ , \ \frac12 \ , \ \frac14 \right) \ \ , \ \ \left( \frac12 \ , \ \frac14 \ , \ \frac14 \right) $ $ \rightarrow \ \ \text{e.g. ,} \ \ f\left( \frac14 \ , \ \frac14 \ , \ \frac12 \right) \ = \ \left(\frac14 \right)^2 \ + \ 2 · \left(\frac14 \right) · \left(\frac12 \right) \ - \ 2 · \left(\frac14 \right)^2 \ · \left(\frac12 \right) \ \ = \ \ \frac14 \ \ ; \ \ $ or

pair of coordinates $ \ = \ \frac12 \ , $ third $ \mathbf{\ = \ 0 \ :} \quad \left( \frac12 \ , \ \frac12 \ , \ 0 \right) \ \ , \ \ \left( \frac12 \ , \ 0 \ , \ \frac12 \right) \ \ , \ \ \left( 0 \ , \ \frac12 \ , \ \frac12 \right) $ $ \rightarrow \ \ \text{e.g. ,} \ \ f\left( \frac14 \ , \ \frac14 \ , \ \frac12 \right) \ = \ \left(\frac12 \right)^2 \ + \ 2 · \left(\frac12 \right) · 0 \ - \ 2 · 0 \ \ = \ \ \frac14 \ \ . $

These cases include points on the intersections of the constraint plane with the coordinate planes, which form an equilateral triangle. Note, as indicated in the graph immediately below, that the vertices of this triangle are not points of tangency with the $ \ f(x,y,z) = 0 \ $ level surface. So the function must be evaluated separately at these "boundary points" $ \ (1 \ , \ 0 \ , \ 0 ) \ , \ (0 \ , \ 1 \ , \ 0 ) \ , \ (0 \ , \ 0 \ , \ 1 ) \ \ ; $ these provide the lower limit for the inequality under discussion. The maximum value $ \ \frac{7}{27} \ $ for the function on the constraint plane then lies at $ \ \left( \frac13 \ , \ \frac13 \ , \ \frac13 \right) \ \ . $ Thus, $ \ 0 \le f(x,y,z) \le \frac{7}{27} \ \ . $

This next graph depicts the level-surface for $ \ f(x,y,z) \ = \ \frac14 \ $ and the constraint plane, with the marked points indicating the points of tangency on the coordinate planes.

Finally, the level-surface $ \ f(x,y,z) \ = \ \frac{7}{27} \ $ and the constraint plane are shown, with the single tangent point $ \ \left( \frac13 \ , \ \frac13 \ , \ \frac13 \right) \ \ . $ [The value of the function is reduced very slightly here to make that point more easily visible.]