Mixing
Implication of equations
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Just an interesting question I saw online. :)
Is the following statement true or false?
$$large dfracabc^2 + dfracbca^2 + dfraccab^2 = 3 iff dfraca + bc + dfracb + ca + dfracc + ab in -3; 6$$
$a + b + c ne 0$
Edit: In case you don't know, $a, b, c$ can be negative numbers too.
algebra-precalculus elementary-number-theory
asked 5 hours ago
Vee Hua Zhi
74718
add a comment |Â
up vote
5
down vote
favorite
Just an interesting question I saw online. :)
Is the following statement true or false?
$$large dfracabc^2 + dfracbca^2 + dfraccab^2 = 3 iff dfraca + bc + dfracb + ca + dfracc + ab in -3; 6$$
$a + b + c ne 0$
Edit: In case you don't know, $a, b, c$ can be negative numbers too.
algebra-precalculus elementary-number-theory
asked 5 hours ago
Vee Hua Zhi
74718
Is here assumed to be $$a,b,c>0$$?
– Dr. Sonnhard Graubner
5 hours ago
@Dr.SonnhardGraubner No. :) Thanks for your interest
– Vee Hua Zhi
5 hours ago
I take it as a know because on the right-hand side there exists a $-3$ and a negative number cannot be formed by additions of positive numbers.
– Vee Hua Zhi
5 hours ago
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Just an interesting question I saw online. :)
Is the following statement true or false?
$$large dfracabc^2 + dfracbca^2 + dfraccab^2 = 3 iff dfraca + bc + dfracb + ca + dfracc + ab in -3; 6$$
$a + b + c ne 0$
Edit: In case you don't know, $a, b, c$ can be negative numbers too.
algebra-precalculus elementary-number-theory
asked 5 hours ago
Vee Hua Zhi
74718
Just an interesting question I saw online. :)
Is the following statement true or false?
$$large dfracabc^2 + dfracbca^2 + dfraccab^2 = 3 iff dfraca + bc + dfracb + ca + dfracc + ab in -3; 6$$
$a + b + c ne 0$
Edit: In case you don't know, $a, b, c$ can be negative numbers too.
algebra-precalculus elementary-number-theory
algebra-precalculus elementary-number-theory
asked 5 hours ago
Vee Hua Zhi
74718
asked 5 hours ago
Vee Hua Zhi
74718
edited 5 hours ago
asked 5 hours ago
Vee Hua Zhi
74718
asked 5 hours ago
Vee Hua Zhi
74718
asked 5 hours ago
Vee Hua Zhi
74718
74718
Is here assumed to be $$a,b,c>0$$?
– Dr. Sonnhard Graubner
5 hours ago
@Dr.SonnhardGraubner No. :) Thanks for your interest
– Vee Hua Zhi
5 hours ago
I take it as a know because on the right-hand side there exists a $-3$ and a negative number cannot be formed by additions of positive numbers.
– Vee Hua Zhi
5 hours ago
add a comment |Â
Is here assumed to be $$a,b,c>0$$?
– Dr. Sonnhard Graubner
5 hours ago
@Dr.SonnhardGraubner No. :) Thanks for your interest
– Vee Hua Zhi
5 hours ago
I take it as a know because on the right-hand side there exists a $-3$ and a negative number cannot be formed by additions of positive numbers.
– Vee Hua Zhi
5 hours ago
Is here assumed to be $$a,b,c>0$$?
– Dr. Sonnhard Graubner
5 hours ago
Is here assumed to be $$a,b,c>0$$?
– Dr. Sonnhard Graubner
5 hours ago
@Dr.SonnhardGraubner No. :) Thanks for your interest
– Vee Hua Zhi
5 hours ago
@Dr.SonnhardGraubner No. :) Thanks for your interest
– Vee Hua Zhi
5 hours ago
I take it as a know because on the right-hand side there exists a $-3$ and a negative number cannot be formed by additions of positive numbers.
– Vee Hua Zhi
5 hours ago
I take it as a know because on the right-hand side there exists a $-3$ and a negative number cannot be formed by additions of positive numbers.
– Vee Hua Zhi
5 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
Notice that the equality $fracabc+fracbca+fraccab=3$ implies that for $x=ab, y=bc, z=ca$:
$$x^3+y^3+z^3-3xyz=0leftrightarrow (x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)=0$$.
This in turn implies that either $ab+bc+ca=0 hspace0.3cm(1)$ or $a=b=c hspace0.3cm(2)$.
- Case (1):
We easily find by applying (1) solved for $a+b, b+c, c+a $ respectively:
$$
beginalign
&fraca+bc=-fracabc^2\
&fracc+ba=-fracbca^2\
&fraca+cb=-fracacb^2
endalign
$$
and we finally find that
$$fraca+bc+fracc+ba+fraca+cb=-fracabc^2
-fracbca^2
-fracacb^2=-3$$
by the original equation.
Case(2):
when $a=b=c$ it's easy to see that
$$fraca+bc+fracc+ba+fraca+cb=6$$
That means that the values of the algebraic expression in question could be either -3 or 6 given the constraint above.
answered 3 hours ago
DinosaurEgg
3807
add a comment |Â
up vote
1
down vote
If $a,b,c$ are all positive then we have by Am-Gm:
$$dfracabc^2 + dfracbca^2 + dfraccab^2 geq 3sqrt[3]dfracabc^2 cdot dfracbca^2 cdot dfraccab^2 = 3$$
With eqaulity iff $ dfracabc^2 = dfracbca^2 = dfraccab^2$ which is iff $a=b=c$ so the value of expression is $6$.
Now suppose not all are positive. Clearly not all numbers $a,b,c$ can be negative. Suppose $a$ is positive and $b$ negative.
Put $x=a/c$ and $y=b/c$, then we can rewrite starting equation like $$x^3y^3+x^3+y^3=3x^2y^2$$
If we put $m= xy<0$ and $k=x+y$ we get
$$ m^3+k^3-3mk-3m^2=0implies k=-m ;;;rm or ;;;k^2-km+m^2=3m$$
We would like to know what is the value of $$(x+y)(xy+x+y+1)over xy-2$$
i.e. $$E:=k(m+k+1)over m-2$$
If $m=-k$ we get $E=-3$.
If $3m =k^2-km+m^2=k^2+m^2+(k-m)^2over 2geq 0$, then $mgeq 0$ but this is a contradiction since $m<0$.
answered 5 hours ago
greedoid
28.9k93878
But for AM-GM we need $$a>0,b>0,c>0$$
– Dr. Sonnhard Graubner
5 hours ago
Yes, for $n=3$ also
– Dr. Sonnhard Graubner
5 hours ago
and thus a conclusion... can you explain how is $a^2b^2=b^2c^2=c^2a^2$ related to $dfraca + bc + dfracb + ca + dfracc + ab in -3; 6$
– Vee Hua Zhi
5 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Notice that the equality $fracabc+fracbca+fraccab=3$ implies that for $x=ab, y=bc, z=ca$:
$$x^3+y^3+z^3-3xyz=0leftrightarrow (x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)=0$$.
This in turn implies that either $ab+bc+ca=0 hspace0.3cm(1)$ or $a=b=c hspace0.3cm(2)$.
- Case (1):
We easily find by applying (1) solved for $a+b, b+c, c+a $ respectively:
$$
beginalign
&fraca+bc=-fracabc^2\
&fracc+ba=-fracbca^2\
&fraca+cb=-fracacb^2
endalign
$$
and we finally find that
$$fraca+bc+fracc+ba+fraca+cb=-fracabc^2
-fracbca^2
-fracacb^2=-3$$
by the original equation.
Case(2):
when $a=b=c$ it's easy to see that
$$fraca+bc+fracc+ba+fraca+cb=6$$
That means that the values of the algebraic expression in question could be either -3 or 6 given the constraint above.
answered 3 hours ago
DinosaurEgg
3807
add a comment |Â
up vote
3
down vote
Notice that the equality $fracabc+fracbca+fraccab=3$ implies that for $x=ab, y=bc, z=ca$:
$$x^3+y^3+z^3-3xyz=0leftrightarrow (x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)=0$$.
This in turn implies that either $ab+bc+ca=0 hspace0.3cm(1)$ or $a=b=c hspace0.3cm(2)$.
- Case (1):
We easily find by applying (1) solved for $a+b, b+c, c+a $ respectively:
$$
beginalign
&fraca+bc=-fracabc^2\
&fracc+ba=-fracbca^2\
&fraca+cb=-fracacb^2
endalign
$$
and we finally find that
$$fraca+bc+fracc+ba+fraca+cb=-fracabc^2
-fracbca^2
-fracacb^2=-3$$
by the original equation.
Case(2):
when $a=b=c$ it's easy to see that
$$fraca+bc+fracc+ba+fraca+cb=6$$
That means that the values of the algebraic expression in question could be either -3 or 6 given the constraint above.
answered 3 hours ago
DinosaurEgg
3807
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Notice that the equality $fracabc+fracbca+fraccab=3$ implies that for $x=ab, y=bc, z=ca$:
$$x^3+y^3+z^3-3xyz=0leftrightarrow (x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)=0$$.
This in turn implies that either $ab+bc+ca=0 hspace0.3cm(1)$ or $a=b=c hspace0.3cm(2)$.
- Case (1):
We easily find by applying (1) solved for $a+b, b+c, c+a $ respectively:
$$
beginalign
&fraca+bc=-fracabc^2\
&fracc+ba=-fracbca^2\
&fraca+cb=-fracacb^2
endalign
$$
and we finally find that
$$fraca+bc+fracc+ba+fraca+cb=-fracabc^2
-fracbca^2
-fracacb^2=-3$$
by the original equation.
Case(2):
when $a=b=c$ it's easy to see that
$$fraca+bc+fracc+ba+fraca+cb=6$$
That means that the values of the algebraic expression in question could be either -3 or 6 given the constraint above.
answered 3 hours ago
DinosaurEgg
3807
Notice that the equality $fracabc+fracbca+fraccab=3$ implies that for $x=ab, y=bc, z=ca$:
$$x^3+y^3+z^3-3xyz=0leftrightarrow (x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)=0$$.
This in turn implies that either $ab+bc+ca=0 hspace0.3cm(1)$ or $a=b=c hspace0.3cm(2)$.
- Case (1):
We easily find by applying (1) solved for $a+b, b+c, c+a $ respectively:
$$
beginalign
&fraca+bc=-fracabc^2\
&fracc+ba=-fracbca^2\
&fraca+cb=-fracacb^2
endalign
$$
and we finally find that
$$fraca+bc+fracc+ba+fraca+cb=-fracabc^2
-fracbca^2
-fracacb^2=-3$$
by the original equation.
Case(2):
when $a=b=c$ it's easy to see that
$$fraca+bc+fracc+ba+fraca+cb=6$$
That means that the values of the algebraic expression in question could be either -3 or 6 given the constraint above.
answered 3 hours ago
DinosaurEgg
3807
answered 3 hours ago
DinosaurEgg
3807
answered 3 hours ago
DinosaurEgg
3807
answered 3 hours ago
DinosaurEgg
3807
3807
add a comment |Â
add a comment |Â
up vote
1
down vote
If $a,b,c$ are all positive then we have by Am-Gm:
$$dfracabc^2 + dfracbca^2 + dfraccab^2 geq 3sqrt[3]dfracabc^2 cdot dfracbca^2 cdot dfraccab^2 = 3$$
With eqaulity iff $ dfracabc^2 = dfracbca^2 = dfraccab^2$ which is iff $a=b=c$ so the value of expression is $6$.
Now suppose not all are positive. Clearly not all numbers $a,b,c$ can be negative. Suppose $a$ is positive and $b$ negative.
Put $x=a/c$ and $y=b/c$, then we can rewrite starting equation like $$x^3y^3+x^3+y^3=3x^2y^2$$
If we put $m= xy<0$ and $k=x+y$ we get
$$ m^3+k^3-3mk-3m^2=0implies k=-m ;;;rm or ;;;k^2-km+m^2=3m$$
We would like to know what is the value of $$(x+y)(xy+x+y+1)over xy-2$$
i.e. $$E:=k(m+k+1)over m-2$$
If $m=-k$ we get $E=-3$.
If $3m =k^2-km+m^2=k^2+m^2+(k-m)^2over 2geq 0$, then $mgeq 0$ but this is a contradiction since $m<0$.
answered 5 hours ago
greedoid
28.9k93878
But for AM-GM we need $$a>0,b>0,c>0$$
– Dr. Sonnhard Graubner
5 hours ago
Yes, for $n=3$ also
– Dr. Sonnhard Graubner
5 hours ago
and thus a conclusion... can you explain how is $a^2b^2=b^2c^2=c^2a^2$ related to $dfraca + bc + dfracb + ca + dfracc + ab in -3; 6$
– Vee Hua Zhi
5 hours ago
add a comment |Â
up vote
1
down vote
If $a,b,c$ are all positive then we have by Am-Gm:
$$dfracabc^2 + dfracbca^2 + dfraccab^2 geq 3sqrt[3]dfracabc^2 cdot dfracbca^2 cdot dfraccab^2 = 3$$
With eqaulity iff $ dfracabc^2 = dfracbca^2 = dfraccab^2$ which is iff $a=b=c$ so the value of expression is $6$.
Now suppose not all are positive. Clearly not all numbers $a,b,c$ can be negative. Suppose $a$ is positive and $b$ negative.
Put $x=a/c$ and $y=b/c$, then we can rewrite starting equation like $$x^3y^3+x^3+y^3=3x^2y^2$$
If we put $m= xy<0$ and $k=x+y$ we get
$$ m^3+k^3-3mk-3m^2=0implies k=-m ;;;rm or ;;;k^2-km+m^2=3m$$
We would like to know what is the value of $$(x+y)(xy+x+y+1)over xy-2$$
i.e. $$E:=k(m+k+1)over m-2$$
If $m=-k$ we get $E=-3$.
If $3m =k^2-km+m^2=k^2+m^2+(k-m)^2over 2geq 0$, then $mgeq 0$ but this is a contradiction since $m<0$.
answered 5 hours ago
greedoid
28.9k93878
But for AM-GM we need $$a>0,b>0,c>0$$
– Dr. Sonnhard Graubner
5 hours ago
Yes, for $n=3$ also
– Dr. Sonnhard Graubner
5 hours ago
and thus a conclusion... can you explain how is $a^2b^2=b^2c^2=c^2a^2$ related to $dfraca + bc + dfracb + ca + dfracc + ab in -3; 6$
– Vee Hua Zhi
5 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $a,b,c$ are all positive then we have by Am-Gm:
$$dfracabc^2 + dfracbca^2 + dfraccab^2 geq 3sqrt[3]dfracabc^2 cdot dfracbca^2 cdot dfraccab^2 = 3$$
With eqaulity iff $ dfracabc^2 = dfracbca^2 = dfraccab^2$ which is iff $a=b=c$ so the value of expression is $6$.
Now suppose not all are positive. Clearly not all numbers $a,b,c$ can be negative. Suppose $a$ is positive and $b$ negative.
Put $x=a/c$ and $y=b/c$, then we can rewrite starting equation like $$x^3y^3+x^3+y^3=3x^2y^2$$
If we put $m= xy<0$ and $k=x+y$ we get
$$ m^3+k^3-3mk-3m^2=0implies k=-m ;;;rm or ;;;k^2-km+m^2=3m$$
We would like to know what is the value of $$(x+y)(xy+x+y+1)over xy-2$$
i.e. $$E:=k(m+k+1)over m-2$$
If $m=-k$ we get $E=-3$.
If $3m =k^2-km+m^2=k^2+m^2+(k-m)^2over 2geq 0$, then $mgeq 0$ but this is a contradiction since $m<0$.
answered 5 hours ago
greedoid
28.9k93878
If $a,b,c$ are all positive then we have by Am-Gm:
$$dfracabc^2 + dfracbca^2 + dfraccab^2 geq 3sqrt[3]dfracabc^2 cdot dfracbca^2 cdot dfraccab^2 = 3$$
With eqaulity iff $ dfracabc^2 = dfracbca^2 = dfraccab^2$ which is iff $a=b=c$ so the value of expression is $6$.
Now suppose not all are positive. Clearly not all numbers $a,b,c$ can be negative. Suppose $a$ is positive and $b$ negative.
Put $x=a/c$ and $y=b/c$, then we can rewrite starting equation like $$x^3y^3+x^3+y^3=3x^2y^2$$
If we put $m= xy<0$ and $k=x+y$ we get
$$ m^3+k^3-3mk-3m^2=0implies k=-m ;;;rm or ;;;k^2-km+m^2=3m$$
We would like to know what is the value of $$(x+y)(xy+x+y+1)over xy-2$$
i.e. $$E:=k(m+k+1)over m-2$$
If $m=-k$ we get $E=-3$.
If $3m =k^2-km+m^2=k^2+m^2+(k-m)^2over 2geq 0$, then $mgeq 0$ but this is a contradiction since $m<0$.
answered 5 hours ago
greedoid
28.9k93878
edited 4 hours ago
answered 5 hours ago
greedoid
28.9k93878
answered 5 hours ago
greedoid
28.9k93878
answered 5 hours ago
greedoid
28.9k93878
28.9k93878
But for AM-GM we need $$a>0,b>0,c>0$$
– Dr. Sonnhard Graubner
5 hours ago
Yes, for $n=3$ also
– Dr. Sonnhard Graubner
5 hours ago
and thus a conclusion... can you explain how is $a^2b^2=b^2c^2=c^2a^2$ related to $dfraca + bc + dfracb + ca + dfracc + ab in -3; 6$
– Vee Hua Zhi
5 hours ago
add a comment |Â
But for AM-GM we need $$a>0,b>0,c>0$$
– Dr. Sonnhard Graubner
5 hours ago
Yes, for $n=3$ also
– Dr. Sonnhard Graubner
5 hours ago
and thus a conclusion... can you explain how is $a^2b^2=b^2c^2=c^2a^2$ related to $dfraca + bc + dfracb + ca + dfracc + ab in -3; 6$
– Vee Hua Zhi
5 hours ago
But for AM-GM we need $$a>0,b>0,c>0$$
– Dr. Sonnhard Graubner
5 hours ago
But for AM-GM we need $$a>0,b>0,c>0$$
– Dr. Sonnhard Graubner
5 hours ago
Yes, for $n=3$ also
– Dr. Sonnhard Graubner
5 hours ago
Yes, for $n=3$ also
– Dr. Sonnhard Graubner
5 hours ago
and thus a conclusion... can you explain how is $a^2b^2=b^2c^2=c^2a^2$ related to $dfraca + bc + dfracb + ca + dfracc + ab in -3; 6$
– Vee Hua Zhi
5 hours ago
and thus a conclusion... can you explain how is $a^2b^2=b^2c^2=c^2a^2$ related to $dfraca + bc + dfracb + ca + dfracc + ab in -3; 6$
– Vee Hua Zhi
5 hours ago
add a comment |Â
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Is here assumed to be $$a,b,c>0$$?
– Dr. Sonnhard Graubner
5 hours ago
@Dr.SonnhardGraubner No. :) Thanks for your interest
– Vee Hua Zhi
5 hours ago
I take it as a know because on the right-hand side there exists a $-3$ and a negative number cannot be formed by additions of positive numbers.
– Vee Hua Zhi
5 hours ago