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What is AM-CM inequality?

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5

0

$begingroup$

There's AM-GM inequality. But I've bumped into AM-CM inequality in official solutions for IMO 2018 problems. Here:

This is A7 problem, solution 1 from here: https://www.imo-official.org/problems/IMO2018SL.pdf

I attempted to use AM-GM inequality in case they simply misspelled, but AM-GM doesn't seem to work here.

Any ideas?

inequality contest-math

share|cite|improve this question

asked 8 hours ago

user75619user75619

39811 silver badge1313 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  CM probably stands for cubic mean.
  $endgroup$
  – Kezer
  8 hours ago
  

  
  
  
  

add a comment | 

5

0

$begingroup$

There's AM-GM inequality. But I've bumped into AM-CM inequality in official solutions for IMO 2018 problems. Here:

This is A7 problem, solution 1 from here: https://www.imo-official.org/problems/IMO2018SL.pdf

I attempted to use AM-GM inequality in case they simply misspelled, but AM-GM doesn't seem to work here.

Any ideas?

inequality contest-math

share|cite|improve this question

asked 8 hours ago

user75619user75619

39811 silver badge1313 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  CM probably stands for cubic mean.
  $endgroup$
  – Kezer
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

There's AM-GM inequality. But I've bumped into AM-CM inequality in official solutions for IMO 2018 problems. Here:

This is A7 problem, solution 1 from here: https://www.imo-official.org/problems/IMO2018SL.pdf

I attempted to use AM-GM inequality in case they simply misspelled, but AM-GM doesn't seem to work here.

Any ideas?

inequality contest-math

share|cite|improve this question

asked 8 hours ago

user75619user75619

39811 silver badge1313 bronze badges

$endgroup$

share|cite|improve this question

asked 8 hours ago

user75619user75619

39811 silver badge1313 bronze badges

share|cite|improve this question

asked 8 hours ago

user75619user75619

39811 silver badge1313 bronze badges

asked 8 hours ago

user75619user75619

39811 silver badge1313 bronze badges

asked 8 hours ago

user75619user75619

39811 silver badge1313 bronze badges

4 Answers
4

active

oldest

votes

4

$begingroup$

The AM-CM inequality likely denotes the Arithmetic mean - Cubic mean inequality which is a specific instance of the Generalized mean inequality. This inequality follows directly from Jensen's inequality mentioned in other answers.

share|cite|improve this answer

answered 8 hours ago

Peter ForemanPeter Foreman

12.5k11 gold badge55 silver badges2828 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

Given what is going on, what they are using is the fact that if a function $f:Itomathbb R$ is convex (here $Isubsetmathbb R$ is an interval), and $x,yin I$, then
$$fleft(fracx+y2right)leqfracf(x)+f(y)2$$
or, equivalently, if $f:Itomathbb R$ is concave, then
$$fracf(x)+f(y)2leq fleft(fracx+y2right).$$
Some people take the statement as the definition of a convex function, and then a function $f$ is concave if $-f$ is convex.

I imagine the "C" in AM-CM stands for either "convex" or "concave".

share|cite|improve this answer

answered 8 hours ago

AweyganAweygan

15.9k22 gold badges1717 silver badges4343 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See also Jensen's inequality.
  $endgroup$
  – Peter Foreman
  8 hours ago
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Elementary approach: no Jensen's inequality or AM-GM inequality (or even AM-CM inequality). BTW I warmly recommend the REAL Shortlist 2018.

It suffices to show that if $a,bgeq 0$ then
$$left(fraca+b2right)^3leq fraca^3+b^32tag1$$
which is equivalent to
$$0leq a^3+b^3-a^2b-ab^2=(a+b)(a-b)^2$$
that trivially holds.

Then apply (1) by letting $a=sqrt[3]fracx+t+147$ and $b=sqrt[3]fracy+t+147$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Robert ZRobert Z

107k1010 gold badges7676 silver badges149149 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the REAL Shortlist! :)
  $endgroup$
  – user75619
  3 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

They appear to be using Jensen's inequality with (Wikipedia's notation) $x_1 = fracx+t+147$, $x_2=fracy+z+147$ and $a_1 = a_2 = 1$ and $varphi(x) = sqrt[3]x$.

share|cite|improve this answer

answered 8 hours ago

ploosu2ploosu2

4,8151010 silver badges2525 bronze badges

$endgroup$

add a comment | 

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4

$begingroup$

The AM-CM inequality likely denotes the Arithmetic mean - Cubic mean inequality which is a specific instance of the Generalized mean inequality. This inequality follows directly from Jensen's inequality mentioned in other answers.

share|cite|improve this answer

answered 8 hours ago

Peter ForemanPeter Foreman

12.5k11 gold badge55 silver badges2828 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

The AM-CM inequality likely denotes the Arithmetic mean - Cubic mean inequality which is a specific instance of the Generalized mean inequality. This inequality follows directly from Jensen's inequality mentioned in other answers.

share|cite|improve this answer

answered 8 hours ago

Peter ForemanPeter Foreman

12.5k11 gold badge55 silver badges2828 bronze badges

$endgroup$

add a comment | 

$begingroup$

The AM-CM inequality likely denotes the Arithmetic mean - Cubic mean inequality which is a specific instance of the Generalized mean inequality. This inequality follows directly from Jensen's inequality mentioned in other answers.

share|cite|improve this answer

answered 8 hours ago

Peter ForemanPeter Foreman

12.5k11 gold badge55 silver badges2828 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Peter ForemanPeter Foreman

12.5k11 gold badge55 silver badges2828 bronze badges

answered 8 hours ago

Peter ForemanPeter Foreman

12.5k11 gold badge55 silver badges2828 bronze badges

answered 8 hours ago

Peter ForemanPeter Foreman

12.5k11 gold badge55 silver badges2828 bronze badges

5

$begingroup$

Given what is going on, what they are using is the fact that if a function $f:Itomathbb R$ is convex (here $Isubsetmathbb R$ is an interval), and $x,yin I$, then
$$fleft(fracx+y2right)leqfracf(x)+f(y)2$$
or, equivalently, if $f:Itomathbb R$ is concave, then
$$fracf(x)+f(y)2leq fleft(fracx+y2right).$$
Some people take the statement as the definition of a convex function, and then a function $f$ is concave if $-f$ is convex.

I imagine the "C" in AM-CM stands for either "convex" or "concave".

share|cite|improve this answer

answered 8 hours ago

AweyganAweygan

15.9k22 gold badges1717 silver badges4343 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See also Jensen's inequality.
  $endgroup$
  – Peter Foreman
  8 hours ago
  
  

  
  
  
  

add a comment | 

5

$begingroup$

Given what is going on, what they are using is the fact that if a function $f:Itomathbb R$ is convex (here $Isubsetmathbb R$ is an interval), and $x,yin I$, then
$$fleft(fracx+y2right)leqfracf(x)+f(y)2$$
or, equivalently, if $f:Itomathbb R$ is concave, then
$$fracf(x)+f(y)2leq fleft(fracx+y2right).$$
Some people take the statement as the definition of a convex function, and then a function $f$ is concave if $-f$ is convex.

I imagine the "C" in AM-CM stands for either "convex" or "concave".

share|cite|improve this answer

answered 8 hours ago

AweyganAweygan

15.9k22 gold badges1717 silver badges4343 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See also Jensen's inequality.
  $endgroup$
  – Peter Foreman
  8 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Given what is going on, what they are using is the fact that if a function $f:Itomathbb R$ is convex (here $Isubsetmathbb R$ is an interval), and $x,yin I$, then
$$fleft(fracx+y2right)leqfracf(x)+f(y)2$$
or, equivalently, if $f:Itomathbb R$ is concave, then
$$fracf(x)+f(y)2leq fleft(fracx+y2right).$$
Some people take the statement as the definition of a convex function, and then a function $f$ is concave if $-f$ is convex.

I imagine the "C" in AM-CM stands for either "convex" or "concave".

share|cite|improve this answer

answered 8 hours ago

AweyganAweygan

15.9k22 gold badges1717 silver badges4343 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

AweyganAweygan

15.9k22 gold badges1717 silver badges4343 bronze badges

answered 8 hours ago

AweyganAweygan

15.9k22 gold badges1717 silver badges4343 bronze badges

answered 8 hours ago

AweyganAweygan

15.9k22 gold badges1717 silver badges4343 bronze badges

1

$begingroup$

Elementary approach: no Jensen's inequality or AM-GM inequality (or even AM-CM inequality). BTW I warmly recommend the REAL Shortlist 2018.

It suffices to show that if $a,bgeq 0$ then
$$left(fraca+b2right)^3leq fraca^3+b^32tag1$$
which is equivalent to
$$0leq a^3+b^3-a^2b-ab^2=(a+b)(a-b)^2$$
that trivially holds.

Then apply (1) by letting $a=sqrt[3]fracx+t+147$ and $b=sqrt[3]fracy+t+147$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Robert ZRobert Z

107k1010 gold badges7676 silver badges149149 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the REAL Shortlist! :)
  $endgroup$
  – user75619
  3 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Elementary approach: no Jensen's inequality or AM-GM inequality (or even AM-CM inequality). BTW I warmly recommend the REAL Shortlist 2018.

It suffices to show that if $a,bgeq 0$ then
$$left(fraca+b2right)^3leq fraca^3+b^32tag1$$
which is equivalent to
$$0leq a^3+b^3-a^2b-ab^2=(a+b)(a-b)^2$$
that trivially holds.

Then apply (1) by letting $a=sqrt[3]fracx+t+147$ and $b=sqrt[3]fracy+t+147$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Robert ZRobert Z

107k1010 gold badges7676 silver badges149149 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the REAL Shortlist! :)
  $endgroup$
  – user75619
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Elementary approach: no Jensen's inequality or AM-GM inequality (or even AM-CM inequality). BTW I warmly recommend the REAL Shortlist 2018.

It suffices to show that if $a,bgeq 0$ then
$$left(fraca+b2right)^3leq fraca^3+b^32tag1$$
which is equivalent to
$$0leq a^3+b^3-a^2b-ab^2=(a+b)(a-b)^2$$
that trivially holds.

Then apply (1) by letting $a=sqrt[3]fracx+t+147$ and $b=sqrt[3]fracy+t+147$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Robert ZRobert Z

107k1010 gold badges7676 silver badges149149 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Robert ZRobert Z

107k1010 gold badges7676 silver badges149149 bronze badges

answered 8 hours ago

Robert ZRobert Z

107k1010 gold badges7676 silver badges149149 bronze badges

answered 8 hours ago

Robert ZRobert Z

107k1010 gold badges7676 silver badges149149 bronze badges

0

$begingroup$

They appear to be using Jensen's inequality with (Wikipedia's notation) $x_1 = fracx+t+147$, $x_2=fracy+z+147$ and $a_1 = a_2 = 1$ and $varphi(x) = sqrt[3]x$.

share|cite|improve this answer

answered 8 hours ago

ploosu2ploosu2

4,8151010 silver badges2525 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

They appear to be using Jensen's inequality with (Wikipedia's notation) $x_1 = fracx+t+147$, $x_2=fracy+z+147$ and $a_1 = a_2 = 1$ and $varphi(x) = sqrt[3]x$.

share|cite|improve this answer

answered 8 hours ago

ploosu2ploosu2

4,8151010 silver badges2525 bronze badges

$endgroup$

add a comment | 

$begingroup$

They appear to be using Jensen's inequality with (Wikipedia's notation) $x_1 = fracx+t+147$, $x_2=fracy+z+147$ and $a_1 = a_2 = 1$ and $varphi(x) = sqrt[3]x$.

share|cite|improve this answer

answered 8 hours ago

ploosu2ploosu2

4,8151010 silver badges2525 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

ploosu2ploosu2

4,8151010 silver badges2525 bronze badges

answered 8 hours ago

ploosu2ploosu2

4,8151010 silver badges2525 bronze badges

answered 8 hours ago

ploosu2ploosu2

4,8151010 silver badges2525 bronze badges