Find the maximum value of the function $M=\sqrt{x^2+y^2-2xy}+\sqrt{y^2+z^2-2yz}+\sqrt{z^2+x^2-2xz}$

Question

Let $0\le x;y;z\le 3$. Find maximum value of function $$M=\sqrt{x^2+y^2-2xy}+\sqrt{y^2+z^2-2yz}+\sqrt{z^2+x^2-2xz}$$

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I can rewrite the function $M=|x-y|+|y-z|+|z-x|$

WLOG $0\le x\le y\le z\le 3$ so we have $M=y-x+z-y+z-x=2z-2x$

I dont know how to use $0\le x;y;z\le 3$ to evaluate the last inequality.

My idea : $M=|x-y|+|y-z|+|z-x|\le \sqrt{((x-y)^2+(y-z)^2+(x-z)^2)(1+1+1)}$

"=" occurs when $|x-y|=|y-z|=|z-x|$ but wrong.

Answer 1

You need to make also one step only:

Let $y=y=0$ and $z=3$.

Thus, we got a value $6$.

We'll prove that it'a maximal value.

Indeed, let $x\geq y\geq z$.

Thus, we need to prove that $$x-y+x-z+y-z\leq6$$ or $$x-z\leq3,$$ which is true because $$x-z\leq x\leq3.$$

Answer 2

You have $M=2z-2x $ as you have done so we need to have $x=0$ in the maximal value of $M$ so $$M\le 2z$$ and $$z\le 3$$ hence $$M\le 6$$

Answer 3

Note that $$ \left| {x - y} \right| + \left| {y - z} \right| + \left| {z - x} \right| = M = const $$ is a cylinder with axis $x=y=z$.

In fact $$ \eqalign{ & M = const = \left| {x - y} \right| + \left| {y - z} \right| + \left| {z - x} \right| = \cr & = \left| {\left( {x - a} \right) - \left( {y - a} \right)} \right| + \left| {\left( {y - a} \right) - \left( {z - a} \right)} \right| + \left| {\left( {z - a} \right) - \left( {x - a} \right)} \right| \cr} $$

The scheme tells you how you can determine the maximum within the given bounds.
For instance we see that, for $y=0$ we get $$ \eqalign{ & \left\{ \matrix{ y = 0 \hfill \cr 0 \le z,x \hfill \cr} \right.\quad \Rightarrow \quad \left| x \right| + \left| z \right| + \left| {z - x} \right| = 2\left| z \right| = M\quad \Rightarrow \cr & \Rightarrow \quad \max M = 6\quad \left| \matrix{ \;0 \le \forall x \le 3 \hfill \cr \;y = 0 \hfill \cr \;z = 3 \hfill \cr} \right. \cr} $$ and the other sets of solutions will be symmetrical to this.

Answer 4

This solution might be a bit far reaching but the idea behind it may also help you in proving many other inequalities.

The following facts give almost immediately the answer to your problem:

• $M(x,y,z) = |x-y|+|y-z|+|z-x|$ is a convex function.
• $Q = \{(x,y,z) \in \mathbb{R}^3\; | \; 0\leq x,y,z \leq 3 \}$ is a compact (bounded and closed) convex set.
• As $M$ is a continuous function on the compact set $Q$, $M$ has a maximum on $Q$.
• Since $M$ and $Q$ are convex, the maximum is attained in an extremal point of $Q$ (the "corners" of the cube $Q$).
• The corners of the cube $Q$ are $C = \{(c_x,c_y,c_z) \in \mathbb{R}^3\; | \; c_x,c_y,c_z \in \{0,3\} \} $

By checking the values of $M$ at the corners you find the maximum at the corners $(3,0,0), (0,3,0), (0,0,3)$: $$\max_Q M = \max_C M = 6$$