How to find the minimum value of $$\frac{x}{3y^2+3z^2+3yz+1}+\frac{y}{3x^2+3z^2+3xz+1}+\frac{z}{3x^2+3y^2+3xy+1}$$,where $x,y,z\geq 0$ and $x+y+z=1$.

It seems to be hard if we use calculus methods. Are there another method? I have no idea. Thank you.


This is again a question of the symmetric type, such as listed in:

With a constraint $\;x+y+z=1\;$ and $\;x,y,z > 0$ . Sort of a general method to transform such a constraint into the inside of a triangle in 2-D has been explained at length in:

The above preliminaries are completed with:

From the latter article comes the following
Theorem (The Purkiss Principle). Let $f$ and $g$ be symmetric functions with continuous second derivatives in the neighborhood of a point $P = (r, \cdots, r)$. On the set where $g$ equals $g(P)$, the function $f$ will have a local maximum or minimum at $P$ except in degenerate cases.

enter image description here

  • Picture on the left: geometry of the conditions $\;x,y,z > 0\;$ and $\;x+y+z=1$ .
  • Picture on the right: contour lines of $f(x,y,z)$ , as seen in the plane of the $\color{red}{red}$ triangle, are at $20$ equidistant levels, between the minimum and the maximum as found within the viewport
Our function $f$ in this case is: $$ f(x,y,z) = \frac{x}{3y^2+3z^2+3yz+1}+\frac{y}{3x^2+3z^2+3xz+1}+\frac{z}{3x^2+3y^2+3xy+1} $$ Applying the Purkiss Principle gives: $$ g(r,r,r)=r+r+r=1 \quad \Longrightarrow \quad r=\frac{1}{3} \quad \Longrightarrow \quad f(r,r,r) = \frac{1}{2} $$ The main problem with the Purkiss Principle, most of the time, is to prove that the extreme found is global. Or at least global enough , i.e. an absolute extreme inside our ($\color{red}{red}$) triangle. In our case, the function will become less than $1/2$, namely close to zero, for $\;|x|,|y|,|z|\rightarrow \infty$ . So it's clear that the Purkiss Principle is "violated", outside the triangle at least. These regions, where $\;f(x,y,z)\;$ is less than, say, $1/2 + 0.001$ , are colored $\color{blue}{blue}$ in the picture on the right.
The $\color{blue}{blue\, spot}$ in the picture on the right is a proof without words that the only minimum inside the triangle is (indeed $= 1/2$ ) and at its center $(x,y,z) = (1/3,1/3,1/3)$ . This finishing touch is thus an informal proof (: disclaimer).

Update : a few further details.

  • Function values close to viewport minimum are white
  • Function values close to viewport maximum are black
  • Contour grey values are just the other way around
  • Function values at the corners of the triangle $=1$ (black)
  • Function values at the edge centers of the triangle $=4/7$


To find a minimum of the function I'm going to find a lower bound and show that the lower bound is attained. For that I'm going to use nothing harder than Cauchy-Schwarz inequality.

Let $f(x, y, z)$ denote your function. Using mentioned Cauchy-Schwarz inequality we get: $$f(x, y, z)\cdot\big(x(3y^2+3z^2+3yz+1) + y(3x^2+3z^2+3xz+1) + z(3x^2+3y^2+3xy+1)\big) \\ \geqslant (x+y+z)^2 = 1.$$

So, $$f(x, y, z) \geqslant \frac{1}{3(xy^2 + xz^2 + yx^2 + yz^2 + zx^2 + zy^2 + 3xyz) + x+y+z} \\ =\frac{1}{3(x+y+z)(xy+yz+xz) + 1} = \frac{1}{3(xy+yz+xz) + 1}$$

We want to get an upper bound of $xy + yz + xz$ under the condition $x + y + z = 1$. We can do this, for example: $$\begin{align} xy + yz + xz &\leqslant x^2 + y^2 + z^2 \\ 3(xy + yz + xz) &\leqslant (x + y + z)^2 = 1 \\ xy + yz + xz &\leqslant \frac{1}{3} \end{align}$$

Finally, $f(x, y, z) \geqslant \dfrac{1}{3\cdot\frac{1}{3} + 1} = \dfrac{1}{2}.$ So $\dfrac{1}{2}$ is the lower bound.
Since $\displaystyle f\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2}$, we can see that we've found a point where the lower bound is attained so this is the minimum of the function.


By C-S and AM-GM we obtain: $$\sum_{cyc}\frac{x}{3y^2+3z^2+3yz+1}=\sum_{cyc}\frac{x}{3y^2+3z^2+3yz+(x+y+z)^2}=$$ $$=\sum_{cyc}\frac{x}{x^2+4y^2+4z^2+2xy+2xz+5yz}=$$ $$=\sum_{cyc}\frac{x^2}{x^3+4xy^2+4xz^2+2x^2y+2x^2z+5xyz}\geq$$ $$\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(x^3+6x^2y+6x^2z+5xyz)}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(2x^3+6x^2y+6x^2z+4xyz)}=$$ $$=\frac{(x+y+z)^2}{2(x+y+z)^3}=\frac{1}{2}.$$ The equality occurs for $x=y=z=\frac{1}{3}$, which says that the answer is $\frac{1}{2}$.



$\textbf{Hint:}$ This is not hard using calculus methods, you can use Lagrange Multipliers.

In a nutshell it says that, to find the maximum and minimum value for $f(x,y,z)$ subject to $g(x,y,z)=k$ you find all $x,y,z$ and $\lambda$ such that:

$$\begin{cases} \nabla f(x,y,z) = \lambda \nabla g(x,y,z) \\g(x,y,z) = k \end{cases}$$

Then you evaluate $f$ in such points, the maximum value you get is the maximum value for the function, the minimum value you get is the minimum value for the function.

Remember that $\nabla f(x,y,z) = (f_x,f_y,f_z)$.

Almost always the "tricky" part in problems involving Lagrange Multipliers is to solve the above system, which seems not to be the case in this example.

  • $\begingroup$ Remember you also have the constraints $x,y,z\geq 0$. $\endgroup$ – ploosu2 Feb 18 '14 at 19:32

I used wolfram alpha: http://goo.gl/Cjzigc

enter image description here

Looks like the min is at $(1/3,1/3,1/3)$.

  • $\begingroup$ At $x=y=z=\dfrac{1}{3}$, the value is smaller. $\endgroup$ – Ma Ming Feb 18 '14 at 19:39
  • $\begingroup$ Oops, I took the maxima for minima. Thanks Mark McClure for editing. $\endgroup$ – ploosu2 Feb 18 '14 at 19:51
  • 2
    $\begingroup$ Using Wolfram Alpha is overkill. Symmetry says (c.f. Purkiss Principle) that $x=y=z=1/3\;$ gives the minimum $=1/2$ . $\endgroup$ – Han de Bruijn Feb 18 '14 at 20:50

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