8

If we have real numbers $x, y, z \in [1, 2]$ then what is the maximum of

$$\frac{x^2+y^2+z^2}{xy+xz+yz}$$

I tried to use substitution $x=\frac{3+\sin X}{2}$, $y=\frac{3+\sin Y}{2}$ and $z=\frac{3+\sin Z}{2}$. But the expression became too messy. This is an Olympiad problem (I don't know the source) and I am not allowed to use calculus. I hope someone can provide an insight to this problem!

Michael Rozenberg
  • 203,855
Dedaha
  • 195
  • 1
    Why that substitution? Anyway, the max appears to occur when two of the variables are $1$ and the other is $2$. The expression suggests there's probably a "symmetry" that could be exploited, but nothing jumps out at me. Are you allowed to use calculus? –  Jun 09 '17 at 16:44
  • 1
    I have used similar substitution for easier problems and sometimes worked! But not for this! No it's an Olympiad problem! I am not allowed to use calculus! – Dedaha Jun 09 '17 at 16:48
  • 1
    Please update the question to make it clear that calculus use is NOT allowed – Dashi Jun 09 '17 at 16:51
  • i think it must be $$\frac{x^2+y^2+z^2}{xy+yz+zx}\le \frac{6}{5}$$ – Dr. Sonnhard Graubner Jun 09 '17 at 16:58
  • How did you get it? – Dedaha Jun 09 '17 at 17:35
  • 1
    @Dedaha, likely with WolframAlpha. He's known for using it, and I used it and got the same thing. –  Jun 09 '17 at 17:35
  • 1
    If you manage to prove that $f(x,y,z)=\frac{x^2+y^2+z^2}{xy+xz+yz}$ is (midpoint-)convex, then its maximum over the cube $[1,2]^3$ is surely achieved at a vertex. Then it is straightforward to understand which vertex. – Jack D'Aurizio Jun 09 '17 at 17:44
  • Not sure if this helps but I noticed that $(x+y)^2 + (x+z)^2 + (y+z)^2 = 2(x^2 + y^2 + z^2 + xy + xz + yz)$ – David Jun 09 '17 at 17:52

1 Answers1

5

Let $k$ is a maximal value and $f(x,y,z)=x^2+y^2+z^2-k(xy+xz+yz)$.

Since, $f$ is a convex function of $x$, of $y$ and of $z$, we obtain: $$0=\max_{\{x,y,z\}\subset[1,2]}f=\max_{\{x,y,z\}\subset\{1,2\}}f$$, which for $x=y=1$ and $z=2$ gives $k=\frac{6}{5}$ and we are done!

Michael Rozenberg
  • 203,855
  • Nice idea, but isn't the convexity of $f$ depending on the positive-definiteness of the matrix $$\begin{pmatrix}1 & -k/2 & -k/2 \ -k/2 & 1 & -k/2 \ -k/2 & -k/2 & 1 \end{pmatrix}$$ ? That puts some restrictions on $k$ for your approach to work. – Jack D'Aurizio Jun 09 '17 at 17:56
  • This is heavy for me! – Dedaha Jun 09 '17 at 18:06
  • @Jack D'Aurizio I don't say that $f$ is a convex function. I say that $f$ is a convex function of $x$, of $y$ and of $z$. Which says that $f$ gets a maximal value for an extremal value of $x$, of $y$ and of $z$, which happens for ${x,y,z}={1,2}$. – Michael Rozenberg Jun 09 '17 at 18:07
  • @Dedaha See my answer to Jack D'Aurizio. Ask your question. I am ready to help. – Michael Rozenberg Jun 09 '17 at 18:09
  • @MichaelRozenberg: oh, coordinate-wise convex, that is subtle. (+1) – Jack D'Aurizio Jun 09 '17 at 18:11
  • @MichaelRozenberg why $f$ is a convex function? – Dedaha Jun 09 '17 at 18:12
  • @Dedaha: it is not granted to be a convex function tout court, but for fixed values of $y$ and $x$ we have that $f(x,y,z)$ is a convex function of the $x$-variable and so on. – Jack D'Aurizio Jun 09 '17 at 18:13
  • 1
    @Dedaha $f$ is a convex function of $x$ because (for example) $\frac{\partial^2f}{\partial x^2}=2>0$. – Michael Rozenberg Jun 09 '17 at 18:18
  • @Dedaha when fixing $y,z$, $f$ will become a positive multiple of $$\frac{x^2+a}{x+b}=\frac{(x+b)^2-2b(x+b)+b^2+a}{x+b}=(x+b)+\frac{b^2+a}{x+b}-2b$$ for some $a,b>0$. Since $x\in[1,2]$, it is clear $f$ is convex in $x$. – Vim Jun 10 '17 at 05:54
  • @MichaelRozenberg Why maximum of $f$ is $0$? – Dedaha Jun 10 '17 at 06:20
  • 1
    @Dedaha Because the maximal value, which you want is $k$. In another words, because $\frac{x^2+y^2+z^2}{xy+xz+yz}\leq k$, where $k$ is a maximal value. – Michael Rozenberg Jun 10 '17 at 06:22