Existence of a “scaling factor” in a inequality
$begingroup$
We're given the real numbers $a,b,c$ that satisfy the condition $ale b le c$.
Consider the case where $ble frac{a+b+c}{3}$, then we have $frac{a+b+c}{3} le frac{a+c}{2} le c$ and similarly $frac{a+b+c}{3} le frac{b+c}{2} le c$, Then there exist $lambda ,mu in [0,1]$ such that: $$frac{c+a}{2}=lambda c+(1-lambda)left(frac{a+b+c}{3}right)$$
and $$frac{b+c}{2}=mu c+(1-mu)left(frac{a+b+c}{3}right)$$
I can "intuitively" see why this should be true, but why do such $lambda$ and $mu$ must exist?
elementary-number-theory inequality
asked Jan 3 at 13:13
Spasoje DurovicSpasoje Durovic
38210
$endgroup$
add a comment |
$begingroup$
We're given the real numbers $a,b,c$ that satisfy the condition $ale b le c$.
Consider the case where $ble frac{a+b+c}{3}$, then we have $frac{a+b+c}{3} le frac{a+c}{2} le c$ and similarly $frac{a+b+c}{3} le frac{b+c}{2} le c$, Then there exist $lambda ,mu in [0,1]$ such that: $$frac{c+a}{2}=lambda c+(1-lambda)left(frac{a+b+c}{3}right)$$
and $$frac{b+c}{2}=mu c+(1-mu)left(frac{a+b+c}{3}right)$$
I can "intuitively" see why this should be true, but why do such $lambda$ and $mu$ must exist?
elementary-number-theory inequality
asked Jan 3 at 13:13
Spasoje DurovicSpasoje Durovic
38210
$endgroup$
add a comment |
$begingroup$
We're given the real numbers $a,b,c$ that satisfy the condition $ale b le c$.
Consider the case where $ble frac{a+b+c}{3}$, then we have $frac{a+b+c}{3} le frac{a+c}{2} le c$ and similarly $frac{a+b+c}{3} le frac{b+c}{2} le c$, Then there exist $lambda ,mu in [0,1]$ such that: $$frac{c+a}{2}=lambda c+(1-lambda)left(frac{a+b+c}{3}right)$$
and $$frac{b+c}{2}=mu c+(1-mu)left(frac{a+b+c}{3}right)$$
I can "intuitively" see why this should be true, but why do such $lambda$ and $mu$ must exist?
elementary-number-theory inequality
asked Jan 3 at 13:13
Spasoje DurovicSpasoje Durovic
38210
$endgroup$
We're given the real numbers $a,b,c$ that satisfy the condition $ale b le c$.
Consider the case where $ble frac{a+b+c}{3}$, then we have $frac{a+b+c}{3} le frac{a+c}{2} le c$ and similarly $frac{a+b+c}{3} le frac{b+c}{2} le c$, Then there exist $lambda ,mu in [0,1]$ such that: $$frac{c+a}{2}=lambda c+(1-lambda)left(frac{a+b+c}{3}right)$$
and $$frac{b+c}{2}=mu c+(1-mu)left(frac{a+b+c}{3}right)$$
I can "intuitively" see why this should be true, but why do such $lambda$ and $mu$ must exist?
elementary-number-theory inequality
elementary-number-theory inequality
asked Jan 3 at 13:13
Spasoje DurovicSpasoje Durovic
38210
asked Jan 3 at 13:13
Spasoje DurovicSpasoje Durovic
38210
asked Jan 3 at 13:13
Spasoje DurovicSpasoje Durovic
38210
asked Jan 3 at 13:13
Spasoje DurovicSpasoje Durovic
38210
asked Jan 3 at 13:13
Spasoje DurovicSpasoje Durovic
38210
38210
add a comment |
add a comment |
1 Answer
1
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$begingroup$
With $x = frac{a+b+c}{3}, y= frac{a+c}{2}, z= c$ your question becomes:
If $x le y le z$ then there is a $lambda in [0, 1]$ such that
$$ tag{*}
y = lambda x + (1 - lambda) z , .
$$
If $x = z$ then any $lambda in [0, 1]$ will do, otherwise $(*)$ is equivalent
to
$$
lambda = frac{z-y}{z-x} , ,
$$
which satisfies $0 le lambda le 1$.
answered Jan 3 at 13:23
Martin RMartin R
29.7k33558
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
With $x = frac{a+b+c}{3}, y= frac{a+c}{2}, z= c$ your question becomes:
If $x le y le z$ then there is a $lambda in [0, 1]$ such that
$$ tag{*}
y = lambda x + (1 - lambda) z , .
$$
If $x = z$ then any $lambda in [0, 1]$ will do, otherwise $(*)$ is equivalent
to
$$
lambda = frac{z-y}{z-x} , ,
$$
which satisfies $0 le lambda le 1$.
answered Jan 3 at 13:23
Martin RMartin R
29.7k33558
$endgroup$
add a comment |
$begingroup$
With $x = frac{a+b+c}{3}, y= frac{a+c}{2}, z= c$ your question becomes:
If $x le y le z$ then there is a $lambda in [0, 1]$ such that
$$ tag{*}
y = lambda x + (1 - lambda) z , .
$$
If $x = z$ then any $lambda in [0, 1]$ will do, otherwise $(*)$ is equivalent
to
$$
lambda = frac{z-y}{z-x} , ,
$$
which satisfies $0 le lambda le 1$.
answered Jan 3 at 13:23
Martin RMartin R
29.7k33558
$endgroup$
add a comment |
$begingroup$
With $x = frac{a+b+c}{3}, y= frac{a+c}{2}, z= c$ your question becomes:
If $x le y le z$ then there is a $lambda in [0, 1]$ such that
$$ tag{*}
y = lambda x + (1 - lambda) z , .
$$
If $x = z$ then any $lambda in [0, 1]$ will do, otherwise $(*)$ is equivalent
to
$$
lambda = frac{z-y}{z-x} , ,
$$
which satisfies $0 le lambda le 1$.
answered Jan 3 at 13:23
Martin RMartin R
29.7k33558
$endgroup$
With $x = frac{a+b+c}{3}, y= frac{a+c}{2}, z= c$ your question becomes:
If $x le y le z$ then there is a $lambda in [0, 1]$ such that
$$ tag{*}
y = lambda x + (1 - lambda) z , .
$$
If $x = z$ then any $lambda in [0, 1]$ will do, otherwise $(*)$ is equivalent
to
$$
lambda = frac{z-y}{z-x} , ,
$$
which satisfies $0 le lambda le 1$.
answered Jan 3 at 13:23
Martin RMartin R
29.7k33558
answered Jan 3 at 13:23
Martin RMartin R
29.7k33558
answered Jan 3 at 13:23
Martin RMartin R
29.7k33558
answered Jan 3 at 13:23
Martin RMartin R
29.7k33558
29.7k33558
add a comment |
add a comment |
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