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Proof of an inequality using analytic geometry
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If $p,q,r$ are real numbers and $0<p<q<r$, then $$frac pq +frac qr +frac rp >frac qp +frac rq +frac pr$$
Is this a well-known inequality?
My proof of it is based on analytic geometry:
If you plot the points $P=(p,1/p)$, $Q=(q,1/q)$, $R=(r,1/r)$ in this order, you get the vertices of a triangle circumscribed by an equilateral hyperbola $xy=1$ entirely on the first quadrant (as all numbers $p,q,r$ are positive and because a line cannot intercept a conic in three points). The area of this triangle PQR can be given by the following formula: $$frac 12 begin vmatrix p & frac 1p & 1 \ q & frac 1q & 1 \ r & frac 1r & 1 \ end vmatrix$$
And as these coordinates are written anticlockwise in this determinant, it has to be positive:
$$begin vmatrix p & frac 1p & 1 \ q & frac 1q & 1 \ r & frac 1r & 1 \ end vmatrix>0$$
From that we get
$$frac pq +frac qr +frac rp >frac qp +frac rq +frac pr$$
Is this proof correct?
Is there another, simpler proof for this inequality?
algebra-precalculus proof-verification inequality analytic-geometry
edited 49 mins ago
greedoid
28.7k93878
asked 1 hour ago
MrDudulex
907
add a comment |Â
up vote
5
down vote
favorite
If $p,q,r$ are real numbers and $0<p<q<r$, then $$frac pq +frac qr +frac rp >frac qp +frac rq +frac pr$$
Is this a well-known inequality?
My proof of it is based on analytic geometry:
If you plot the points $P=(p,1/p)$, $Q=(q,1/q)$, $R=(r,1/r)$ in this order, you get the vertices of a triangle circumscribed by an equilateral hyperbola $xy=1$ entirely on the first quadrant (as all numbers $p,q,r$ are positive and because a line cannot intercept a conic in three points). The area of this triangle PQR can be given by the following formula: $$frac 12 begin vmatrix p & frac 1p & 1 \ q & frac 1q & 1 \ r & frac 1r & 1 \ end vmatrix$$
And as these coordinates are written anticlockwise in this determinant, it has to be positive:
$$begin vmatrix p & frac 1p & 1 \ q & frac 1q & 1 \ r & frac 1r & 1 \ end vmatrix>0$$
From that we get
$$frac pq +frac qr +frac rp >frac qp +frac rq +frac pr$$
Is this proof correct?
Is there another, simpler proof for this inequality?
algebra-precalculus proof-verification inequality analytic-geometry
edited 49 mins ago
greedoid
28.7k93878
asked 1 hour ago
MrDudulex
907
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
If $p,q,r$ are real numbers and $0<p<q<r$, then $$frac pq +frac qr +frac rp >frac qp +frac rq +frac pr$$
Is this a well-known inequality?
My proof of it is based on analytic geometry:
If you plot the points $P=(p,1/p)$, $Q=(q,1/q)$, $R=(r,1/r)$ in this order, you get the vertices of a triangle circumscribed by an equilateral hyperbola $xy=1$ entirely on the first quadrant (as all numbers $p,q,r$ are positive and because a line cannot intercept a conic in three points). The area of this triangle PQR can be given by the following formula: $$frac 12 begin vmatrix p & frac 1p & 1 \ q & frac 1q & 1 \ r & frac 1r & 1 \ end vmatrix$$
And as these coordinates are written anticlockwise in this determinant, it has to be positive:
$$begin vmatrix p & frac 1p & 1 \ q & frac 1q & 1 \ r & frac 1r & 1 \ end vmatrix>0$$
From that we get
$$frac pq +frac qr +frac rp >frac qp +frac rq +frac pr$$
Is this proof correct?
Is there another, simpler proof for this inequality?
algebra-precalculus proof-verification inequality analytic-geometry
edited 49 mins ago
greedoid
28.7k93878
asked 1 hour ago
MrDudulex
907
If $p,q,r$ are real numbers and $0<p<q<r$, then $$frac pq +frac qr +frac rp >frac qp +frac rq +frac pr$$
Is this a well-known inequality?
My proof of it is based on analytic geometry:
If you plot the points $P=(p,1/p)$, $Q=(q,1/q)$, $R=(r,1/r)$ in this order, you get the vertices of a triangle circumscribed by an equilateral hyperbola $xy=1$ entirely on the first quadrant (as all numbers $p,q,r$ are positive and because a line cannot intercept a conic in three points). The area of this triangle PQR can be given by the following formula: $$frac 12 begin vmatrix p & frac 1p & 1 \ q & frac 1q & 1 \ r & frac 1r & 1 \ end vmatrix$$
And as these coordinates are written anticlockwise in this determinant, it has to be positive:
$$begin vmatrix p & frac 1p & 1 \ q & frac 1q & 1 \ r & frac 1r & 1 \ end vmatrix>0$$
From that we get
$$frac pq +frac qr +frac rp >frac qp +frac rq +frac pr$$
Is this proof correct?
Is there another, simpler proof for this inequality?
algebra-precalculus proof-verification inequality analytic-geometry
algebra-precalculus proof-verification inequality analytic-geometry
edited 49 mins ago
greedoid
28.7k93878
asked 1 hour ago
MrDudulex
907
edited 49 mins ago
greedoid
28.7k93878
asked 1 hour ago
MrDudulex
907
edited 49 mins ago
greedoid
28.7k93878
edited 49 mins ago
greedoid
28.7k93878
edited 49 mins ago
greedoid
28.7k93878
28.7k93878
asked 1 hour ago
MrDudulex
907
asked 1 hour ago
MrDudulex
907
asked 1 hour ago
MrDudulex
907
907
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
accepted
Just a slightly different view at your proof: If the function $f: I to Bbb R$ is
strictly convex on the interval $I subset Bbb R$ then for $p < q < r$ in $I$
$$
f(q) < fracr-qr-p , f(p) + fracq-pr-p , f(r) \
iff (r-q) , f(p) + (p-r) , f(q) + (q-p) , f(r) > 0 quad (*)
$$
Choosing $f(x) = frac 1x$ gives the desired inequality.
The connection to your solution is that $(*)$ can be written as
$$
begin vmatrix p & f(p) & 1 \ q & f(q) & 1 \ r & f(r) & 1 \ end vmatrix>0
, ,
$$
i.e. the (oriented) area of the triangle is positive because
$f(x) = frac 1x$ is strictly convex.
answered 21 mins ago
Martin R
24.1k32744
Nice fact. Using this fact we can create many homologous inequalities with such convex functions
– MrDudulex
5 mins ago
add a comment |Â
up vote
3
down vote
Put $x=rover q>1$ and $y= qover p>1$ so we have to prove $$1over x+1over y +xy >x+y+1over xy$$ or $$x+y+x^2y^2>x^2y+y^2x+1$$
or $$x^2y^2-xy(x+y)+(x+y)-1>0$$
or $$(xy-1)(xy+1)-(x+y)(xy-1)>0$$ or $$(xy-1)(xy-x-y+1)>0$$
or $$(xy-1)(x-1)(y-1)>0$$
which is true.
answered 1 hour ago
greedoid
28.7k93878
add a comment |Â
up vote
0
down vote
Your inequalitiy is equavalent to $$- left( q-r right) left( p-r right) left( p-q right) >0$$
and this is true since we have $$0<p<q<r$$
answered 1 hour ago
Dr. Sonnhard Graubner
68.9k32760
add a comment |Â
up vote
0
down vote
As an alternative let
$$x=frac pq quad y=frac qr quad frac rp=frac1xy quad x,yin(0,1) $$
therefore we obtain
$$frac pq +frac qr +frac rp >frac qp +frac rq +frac pr iff x+y+frac1xyge frac1x+frac1y+xy$$
$$x+y+frac1xy-frac1x-frac1y-xyge 0iff fracx^2y+xy^2+1-y-x-1xyge 0$$
$$x^2y+xy^2-y-xge 0 iff xy(x+y)-(x+y)ge 0iff (xy-1)(x+y)ge 0$$
which is true.
answered 2 mins ago
gimusi
73.6k73889
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Just a slightly different view at your proof: If the function $f: I to Bbb R$ is
strictly convex on the interval $I subset Bbb R$ then for $p < q < r$ in $I$
$$
f(q) < fracr-qr-p , f(p) + fracq-pr-p , f(r) \
iff (r-q) , f(p) + (p-r) , f(q) + (q-p) , f(r) > 0 quad (*)
$$
Choosing $f(x) = frac 1x$ gives the desired inequality.
The connection to your solution is that $(*)$ can be written as
$$
begin vmatrix p & f(p) & 1 \ q & f(q) & 1 \ r & f(r) & 1 \ end vmatrix>0
, ,
$$
i.e. the (oriented) area of the triangle is positive because
$f(x) = frac 1x$ is strictly convex.
answered 21 mins ago
Martin R
24.1k32744
Nice fact. Using this fact we can create many homologous inequalities with such convex functions
– MrDudulex
5 mins ago
add a comment |Â
up vote
1
down vote
accepted
Just a slightly different view at your proof: If the function $f: I to Bbb R$ is
strictly convex on the interval $I subset Bbb R$ then for $p < q < r$ in $I$
$$
f(q) < fracr-qr-p , f(p) + fracq-pr-p , f(r) \
iff (r-q) , f(p) + (p-r) , f(q) + (q-p) , f(r) > 0 quad (*)
$$
Choosing $f(x) = frac 1x$ gives the desired inequality.
The connection to your solution is that $(*)$ can be written as
$$
begin vmatrix p & f(p) & 1 \ q & f(q) & 1 \ r & f(r) & 1 \ end vmatrix>0
, ,
$$
i.e. the (oriented) area of the triangle is positive because
$f(x) = frac 1x$ is strictly convex.
answered 21 mins ago
Martin R
24.1k32744
Nice fact. Using this fact we can create many homologous inequalities with such convex functions
– MrDudulex
5 mins ago
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Just a slightly different view at your proof: If the function $f: I to Bbb R$ is
strictly convex on the interval $I subset Bbb R$ then for $p < q < r$ in $I$
$$
f(q) < fracr-qr-p , f(p) + fracq-pr-p , f(r) \
iff (r-q) , f(p) + (p-r) , f(q) + (q-p) , f(r) > 0 quad (*)
$$
Choosing $f(x) = frac 1x$ gives the desired inequality.
The connection to your solution is that $(*)$ can be written as
$$
begin vmatrix p & f(p) & 1 \ q & f(q) & 1 \ r & f(r) & 1 \ end vmatrix>0
, ,
$$
i.e. the (oriented) area of the triangle is positive because
$f(x) = frac 1x$ is strictly convex.
answered 21 mins ago
Martin R
24.1k32744
Just a slightly different view at your proof: If the function $f: I to Bbb R$ is
strictly convex on the interval $I subset Bbb R$ then for $p < q < r$ in $I$
$$
f(q) < fracr-qr-p , f(p) + fracq-pr-p , f(r) \
iff (r-q) , f(p) + (p-r) , f(q) + (q-p) , f(r) > 0 quad (*)
$$
Choosing $f(x) = frac 1x$ gives the desired inequality.
The connection to your solution is that $(*)$ can be written as
$$
begin vmatrix p & f(p) & 1 \ q & f(q) & 1 \ r & f(r) & 1 \ end vmatrix>0
, ,
$$
i.e. the (oriented) area of the triangle is positive because
$f(x) = frac 1x$ is strictly convex.
answered 21 mins ago
Martin R
24.1k32744
edited 16 mins ago
answered 21 mins ago
Martin R
24.1k32744
answered 21 mins ago
Martin R
24.1k32744
answered 21 mins ago
Martin R
24.1k32744
24.1k32744
Nice fact. Using this fact we can create many homologous inequalities with such convex functions
– MrDudulex
5 mins ago
add a comment |Â
Nice fact. Using this fact we can create many homologous inequalities with such convex functions
– MrDudulex
5 mins ago
Nice fact. Using this fact we can create many homologous inequalities with such convex functions
– MrDudulex
5 mins ago
Nice fact. Using this fact we can create many homologous inequalities with such convex functions
– MrDudulex
5 mins ago
add a comment |Â
up vote
3
down vote
Put $x=rover q>1$ and $y= qover p>1$ so we have to prove $$1over x+1over y +xy >x+y+1over xy$$ or $$x+y+x^2y^2>x^2y+y^2x+1$$
or $$x^2y^2-xy(x+y)+(x+y)-1>0$$
or $$(xy-1)(xy+1)-(x+y)(xy-1)>0$$ or $$(xy-1)(xy-x-y+1)>0$$
or $$(xy-1)(x-1)(y-1)>0$$
which is true.
answered 1 hour ago
greedoid
28.7k93878
add a comment |Â
up vote
3
down vote
Put $x=rover q>1$ and $y= qover p>1$ so we have to prove $$1over x+1over y +xy >x+y+1over xy$$ or $$x+y+x^2y^2>x^2y+y^2x+1$$
or $$x^2y^2-xy(x+y)+(x+y)-1>0$$
or $$(xy-1)(xy+1)-(x+y)(xy-1)>0$$ or $$(xy-1)(xy-x-y+1)>0$$
or $$(xy-1)(x-1)(y-1)>0$$
which is true.
answered 1 hour ago
greedoid
28.7k93878
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Put $x=rover q>1$ and $y= qover p>1$ so we have to prove $$1over x+1over y +xy >x+y+1over xy$$ or $$x+y+x^2y^2>x^2y+y^2x+1$$
or $$x^2y^2-xy(x+y)+(x+y)-1>0$$
or $$(xy-1)(xy+1)-(x+y)(xy-1)>0$$ or $$(xy-1)(xy-x-y+1)>0$$
or $$(xy-1)(x-1)(y-1)>0$$
which is true.
answered 1 hour ago
greedoid
28.7k93878
Put $x=rover q>1$ and $y= qover p>1$ so we have to prove $$1over x+1over y +xy >x+y+1over xy$$ or $$x+y+x^2y^2>x^2y+y^2x+1$$
or $$x^2y^2-xy(x+y)+(x+y)-1>0$$
or $$(xy-1)(xy+1)-(x+y)(xy-1)>0$$ or $$(xy-1)(xy-x-y+1)>0$$
or $$(xy-1)(x-1)(y-1)>0$$
which is true.
answered 1 hour ago
greedoid
28.7k93878
edited 1 hour ago
answered 1 hour ago
greedoid
28.7k93878
answered 1 hour ago
greedoid
28.7k93878
answered 1 hour ago
greedoid
28.7k93878
28.7k93878
add a comment |Â
add a comment |Â
up vote
0
down vote
Your inequalitiy is equavalent to $$- left( q-r right) left( p-r right) left( p-q right) >0$$
and this is true since we have $$0<p<q<r$$
answered 1 hour ago
Dr. Sonnhard Graubner
68.9k32760
add a comment |Â
up vote
0
down vote
Your inequalitiy is equavalent to $$- left( q-r right) left( p-r right) left( p-q right) >0$$
and this is true since we have $$0<p<q<r$$
answered 1 hour ago
Dr. Sonnhard Graubner
68.9k32760
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your inequalitiy is equavalent to $$- left( q-r right) left( p-r right) left( p-q right) >0$$
and this is true since we have $$0<p<q<r$$
answered 1 hour ago
Dr. Sonnhard Graubner
68.9k32760
Your inequalitiy is equavalent to $$- left( q-r right) left( p-r right) left( p-q right) >0$$
and this is true since we have $$0<p<q<r$$
answered 1 hour ago
Dr. Sonnhard Graubner
68.9k32760
edited 1 hour ago
answered 1 hour ago
Dr. Sonnhard Graubner
68.9k32760
answered 1 hour ago
Dr. Sonnhard Graubner
68.9k32760
answered 1 hour ago
Dr. Sonnhard Graubner
68.9k32760
68.9k32760
add a comment |Â
add a comment |Â
up vote
0
down vote
As an alternative let
$$x=frac pq quad y=frac qr quad frac rp=frac1xy quad x,yin(0,1) $$
therefore we obtain
$$frac pq +frac qr +frac rp >frac qp +frac rq +frac pr iff x+y+frac1xyge frac1x+frac1y+xy$$
$$x+y+frac1xy-frac1x-frac1y-xyge 0iff fracx^2y+xy^2+1-y-x-1xyge 0$$
$$x^2y+xy^2-y-xge 0 iff xy(x+y)-(x+y)ge 0iff (xy-1)(x+y)ge 0$$
which is true.
answered 2 mins ago
gimusi
73.6k73889
add a comment |Â
up vote
0
down vote
As an alternative let
$$x=frac pq quad y=frac qr quad frac rp=frac1xy quad x,yin(0,1) $$
therefore we obtain
$$frac pq +frac qr +frac rp >frac qp +frac rq +frac pr iff x+y+frac1xyge frac1x+frac1y+xy$$
$$x+y+frac1xy-frac1x-frac1y-xyge 0iff fracx^2y+xy^2+1-y-x-1xyge 0$$
$$x^2y+xy^2-y-xge 0 iff xy(x+y)-(x+y)ge 0iff (xy-1)(x+y)ge 0$$
which is true.
answered 2 mins ago
gimusi
73.6k73889
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As an alternative let
$$x=frac pq quad y=frac qr quad frac rp=frac1xy quad x,yin(0,1) $$
therefore we obtain
$$frac pq +frac qr +frac rp >frac qp +frac rq +frac pr iff x+y+frac1xyge frac1x+frac1y+xy$$
$$x+y+frac1xy-frac1x-frac1y-xyge 0iff fracx^2y+xy^2+1-y-x-1xyge 0$$
$$x^2y+xy^2-y-xge 0 iff xy(x+y)-(x+y)ge 0iff (xy-1)(x+y)ge 0$$
which is true.
answered 2 mins ago
gimusi
73.6k73889
As an alternative let
$$x=frac pq quad y=frac qr quad frac rp=frac1xy quad x,yin(0,1) $$
therefore we obtain
$$frac pq +frac qr +frac rp >frac qp +frac rq +frac pr iff x+y+frac1xyge frac1x+frac1y+xy$$
$$x+y+frac1xy-frac1x-frac1y-xyge 0iff fracx^2y+xy^2+1-y-x-1xyge 0$$
$$x^2y+xy^2-y-xge 0 iff xy(x+y)-(x+y)ge 0iff (xy-1)(x+y)ge 0$$
which is true.
answered 2 mins ago
gimusi
73.6k73889
answered 2 mins ago
gimusi
73.6k73889
answered 2 mins ago
gimusi
73.6k73889
answered 2 mins ago
gimusi
73.6k73889
73.6k73889
add a comment |Â
add a comment |Â
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