Find the maximum value of the function $M=sqrt{x^2+y^2-2xy}+sqrt{y^2+z^2-2yz}+sqrt{z^2+x^2-2xz}$
$begingroup$
Let $0le x;y;zle 3$. Find maximum value of function $$M=sqrt{x^2+y^2-2xy}+sqrt{y^2+z^2-2yz}+sqrt{z^2+x^2-2xz}$$
I can rewrite the function $M=|x-y|+|y-z|+|z-x|$
WLOG $0le xle yle zle 3$ so we have $M=y-x+z-y+z-x=2z-2x$
I dont know how to use $0le x;y;zle 3$ 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.
inequality radicals maxima-minima
edited Dec 24 '18 at 17:11
Michael Rozenberg
107k1894198
asked Dec 22 '18 at 3:50
nDLynknDLynk
20010
$endgroup$
add a comment |
$begingroup$
Let $0le x;y;zle 3$. Find maximum value of function $$M=sqrt{x^2+y^2-2xy}+sqrt{y^2+z^2-2yz}+sqrt{z^2+x^2-2xz}$$
I can rewrite the function $M=|x-y|+|y-z|+|z-x|$
WLOG $0le xle yle zle 3$ so we have $M=y-x+z-y+z-x=2z-2x$
I dont know how to use $0le x;y;zle 3$ 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.
inequality radicals maxima-minima
edited Dec 24 '18 at 17:11
Michael Rozenberg
107k1894198
asked Dec 22 '18 at 3:50
nDLynknDLynk
20010
$endgroup$
$begingroup$
M = 2(z-x) then z=3 and x=0
$endgroup$
– eyllanesc
Dec 22 '18 at 4:28
add a comment |
$begingroup$
Let $0le x;y;zle 3$. Find maximum value of function $$M=sqrt{x^2+y^2-2xy}+sqrt{y^2+z^2-2yz}+sqrt{z^2+x^2-2xz}$$
I can rewrite the function $M=|x-y|+|y-z|+|z-x|$
WLOG $0le xle yle zle 3$ so we have $M=y-x+z-y+z-x=2z-2x$
I dont know how to use $0le x;y;zle 3$ 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.
inequality radicals maxima-minima
edited Dec 24 '18 at 17:11
Michael Rozenberg
107k1894198
asked Dec 22 '18 at 3:50
nDLynknDLynk
20010
$endgroup$
Let $0le x;y;zle 3$. Find maximum value of function $$M=sqrt{x^2+y^2-2xy}+sqrt{y^2+z^2-2yz}+sqrt{z^2+x^2-2xz}$$
I can rewrite the function $M=|x-y|+|y-z|+|z-x|$
WLOG $0le xle yle zle 3$ so we have $M=y-x+z-y+z-x=2z-2x$
I dont know how to use $0le x;y;zle 3$ 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.
inequality radicals maxima-minima
inequality radicals maxima-minima
edited Dec 24 '18 at 17:11
Michael Rozenberg
107k1894198
asked Dec 22 '18 at 3:50
nDLynknDLynk
20010
edited Dec 24 '18 at 17:11
Michael Rozenberg
107k1894198
asked Dec 22 '18 at 3:50
nDLynknDLynk
20010
edited Dec 24 '18 at 17:11
Michael Rozenberg
107k1894198
edited Dec 24 '18 at 17:11
Michael Rozenberg
107k1894198
edited Dec 24 '18 at 17:11
Michael Rozenberg
107k1894198
107k1894198
asked Dec 22 '18 at 3:50
nDLynknDLynk
20010
asked Dec 22 '18 at 3:50
nDLynknDLynk
20010
asked Dec 22 '18 at 3:50
nDLynknDLynk
20010
20010
$begingroup$
M = 2(z-x) then z=3 and x=0
$endgroup$
– eyllanesc
Dec 22 '18 at 4:28
add a comment |
$begingroup$
M = 2(z-x) then z=3 and x=0
$endgroup$
– eyllanesc
Dec 22 '18 at 4:28
$begingroup$
M = 2(z-x) then z=3 and x=0
$endgroup$
– eyllanesc
Dec 22 '18 at 4:28
$begingroup$
M = 2(z-x) then z=3 and x=0
$endgroup$
– eyllanesc
Dec 22 '18 at 4:28
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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 $xgeq ygeq z$.
Thus, we need to prove that
$$x-y+x-z+y-zleq6$$ or $$x-zleq3,$$ which is true because
$$x-zleq xleq3.$$
answered Dec 22 '18 at 5:31
Michael RozenbergMichael Rozenberg
107k1894198
$endgroup$
add a comment |
$begingroup$
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
$$
eqalign{
& left{ matrix{
y = 0 hfill cr
0 le z,x hfill cr} right.quad Rightarrow quad left| x right| + left| z right| + left| {z - x} right| = 2left| z right| = Mquad Rightarrow cr
& Rightarrow quad max M = 6quad left| matrix{
;0 le forall x le 3 hfill cr
;y = 0 hfill cr
;z = 3 hfill cr} right. cr}
$$
and the other sets of solutions will be symmetrical to this.
answered Dec 22 '18 at 18:12
G CabG Cab
19.9k31340
$endgroup$
add a comment |
$begingroup$
You have $M=2z-2x $ as you have done
so we need to have $x=0$ in the maximal value of $M$
so $$Mle 2z$$
and $$zle 3$$
hence $$Mle 6$$
answered Dec 22 '18 at 16:24
user600785user600785
6810
$endgroup$
add a comment |
$begingroup$
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.
$Q = {(x,y,z) in mathbb{R}^3; | ; 0leq x,y,z leq 3 }$ 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 $(3,0,0), (0,3,0), (0,0,3)$:
$$max_Q M = max_C M = 6$$
answered Dec 23 '18 at 4:00
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $xgeq ygeq z$.
Thus, we need to prove that
$$x-y+x-z+y-zleq6$$ or $$x-zleq3,$$ which is true because
$$x-zleq xleq3.$$
answered Dec 22 '18 at 5:31
Michael RozenbergMichael Rozenberg
107k1894198
$endgroup$
add a comment |
$begingroup$
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 $xgeq ygeq z$.
Thus, we need to prove that
$$x-y+x-z+y-zleq6$$ or $$x-zleq3,$$ which is true because
$$x-zleq xleq3.$$
answered Dec 22 '18 at 5:31
Michael RozenbergMichael Rozenberg
107k1894198
$endgroup$
add a comment |
$begingroup$
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 $xgeq ygeq z$.
Thus, we need to prove that
$$x-y+x-z+y-zleq6$$ or $$x-zleq3,$$ which is true because
$$x-zleq xleq3.$$
answered Dec 22 '18 at 5:31
Michael RozenbergMichael Rozenberg
107k1894198
$endgroup$
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 $xgeq ygeq z$.
Thus, we need to prove that
$$x-y+x-z+y-zleq6$$ or $$x-zleq3,$$ which is true because
$$x-zleq xleq3.$$
answered Dec 22 '18 at 5:31
Michael RozenbergMichael Rozenberg
107k1894198
answered Dec 22 '18 at 5:31
Michael RozenbergMichael Rozenberg
107k1894198
answered Dec 22 '18 at 5:31
Michael RozenbergMichael Rozenberg
107k1894198
answered Dec 22 '18 at 5:31
Michael RozenbergMichael Rozenberg
107k1894198
107k1894198
add a comment |
add a comment |
$begingroup$
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
$$
eqalign{
& left{ matrix{
y = 0 hfill cr
0 le z,x hfill cr} right.quad Rightarrow quad left| x right| + left| z right| + left| {z - x} right| = 2left| z right| = Mquad Rightarrow cr
& Rightarrow quad max M = 6quad left| matrix{
;0 le forall x le 3 hfill cr
;y = 0 hfill cr
;z = 3 hfill cr} right. cr}
$$
and the other sets of solutions will be symmetrical to this.
answered Dec 22 '18 at 18:12
G CabG Cab
19.9k31340
$endgroup$
add a comment |
$begingroup$
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
$$
eqalign{
& left{ matrix{
y = 0 hfill cr
0 le z,x hfill cr} right.quad Rightarrow quad left| x right| + left| z right| + left| {z - x} right| = 2left| z right| = Mquad Rightarrow cr
& Rightarrow quad max M = 6quad left| matrix{
;0 le forall x le 3 hfill cr
;y = 0 hfill cr
;z = 3 hfill cr} right. cr}
$$
and the other sets of solutions will be symmetrical to this.
answered Dec 22 '18 at 18:12
G CabG Cab
19.9k31340
$endgroup$
add a comment |
$begingroup$
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
$$
eqalign{
& left{ matrix{
y = 0 hfill cr
0 le z,x hfill cr} right.quad Rightarrow quad left| x right| + left| z right| + left| {z - x} right| = 2left| z right| = Mquad Rightarrow cr
& Rightarrow quad max M = 6quad left| matrix{
;0 le forall x le 3 hfill cr
;y = 0 hfill cr
;z = 3 hfill cr} right. cr}
$$
and the other sets of solutions will be symmetrical to this.
answered Dec 22 '18 at 18:12
G CabG Cab
19.9k31340
$endgroup$
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
$$
eqalign{
& left{ matrix{
y = 0 hfill cr
0 le z,x hfill cr} right.quad Rightarrow quad left| x right| + left| z right| + left| {z - x} right| = 2left| z right| = Mquad Rightarrow cr
& Rightarrow quad max M = 6quad left| matrix{
;0 le forall x le 3 hfill cr
;y = 0 hfill cr
;z = 3 hfill cr} right. cr}
$$
and the other sets of solutions will be symmetrical to this.
answered Dec 22 '18 at 18:12
G CabG Cab
19.9k31340
answered Dec 22 '18 at 18:12
G CabG Cab
19.9k31340
answered Dec 22 '18 at 18:12
G CabG Cab
19.9k31340
answered Dec 22 '18 at 18:12
G CabG Cab
19.9k31340
19.9k31340
add a comment |
add a comment |
$begingroup$
You have $M=2z-2x $ as you have done
so we need to have $x=0$ in the maximal value of $M$
so $$Mle 2z$$
and $$zle 3$$
hence $$Mle 6$$
answered Dec 22 '18 at 16:24
user600785user600785
6810
$endgroup$
add a comment |
$begingroup$
You have $M=2z-2x $ as you have done
so we need to have $x=0$ in the maximal value of $M$
so $$Mle 2z$$
and $$zle 3$$
hence $$Mle 6$$
answered Dec 22 '18 at 16:24
user600785user600785
6810
$endgroup$
add a comment |
$begingroup$
You have $M=2z-2x $ as you have done
so we need to have $x=0$ in the maximal value of $M$
so $$Mle 2z$$
and $$zle 3$$
hence $$Mle 6$$
answered Dec 22 '18 at 16:24
user600785user600785
6810
$endgroup$
You have $M=2z-2x $ as you have done
so we need to have $x=0$ in the maximal value of $M$
so $$Mle 2z$$
and $$zle 3$$
hence $$Mle 6$$
answered Dec 22 '18 at 16:24
user600785user600785
6810
answered Dec 22 '18 at 16:24
user600785user600785
6810
answered Dec 22 '18 at 16:24
user600785user600785
6810
answered Dec 22 '18 at 16:24
user600785user600785
6810
6810
add a comment |
add a comment |
$begingroup$
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.
$Q = {(x,y,z) in mathbb{R}^3; | ; 0leq x,y,z leq 3 }$ 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 $(3,0,0), (0,3,0), (0,0,3)$:
$$max_Q M = max_C M = 6$$
answered Dec 23 '18 at 4:00
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
$begingroup$
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.
$Q = {(x,y,z) in mathbb{R}^3; | ; 0leq x,y,z leq 3 }$ 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 $(3,0,0), (0,3,0), (0,0,3)$:
$$max_Q M = max_C M = 6$$
answered Dec 23 '18 at 4:00
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
$begingroup$
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.
$Q = {(x,y,z) in mathbb{R}^3; | ; 0leq x,y,z leq 3 }$ 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 $(3,0,0), (0,3,0), (0,0,3)$:
$$max_Q M = max_C M = 6$$
answered Dec 23 '18 at 4:00
trancelocationtrancelocation
12.6k1826
$endgroup$
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.
$Q = {(x,y,z) in mathbb{R}^3; | ; 0leq x,y,z leq 3 }$ 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 $(3,0,0), (0,3,0), (0,0,3)$:
$$max_Q M = max_C M = 6$$
answered Dec 23 '18 at 4:00
trancelocationtrancelocation
12.6k1826
answered Dec 23 '18 at 4:00
trancelocationtrancelocation
12.6k1826
answered Dec 23 '18 at 4:00
trancelocationtrancelocation
12.6k1826
answered Dec 23 '18 at 4:00
trancelocationtrancelocation
12.6k1826
12.6k1826
add a comment |
add a comment |
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$begingroup$
M = 2(z-x) then z=3 and x=0
$endgroup$
– eyllanesc
Dec 22 '18 at 4:28