Inequalities - Tangent line trick
I will first state the "trick":
we fix $a=frac{a_1+a_2+...+a_n}{n}, $If $f$ is not convex we can sometimes prove:$$f(x)ge f(a)+f'(a)(x-a) $$ If this manages to hold for all x, then summing up the inequality will give us the desired conclusion.
Then we have the following example:
We want to show, given that $a+b+c=3$: $$sum_{cyc}(frac{18}{(3-c)(4-c)}-c^2) ge 6 $$Using the tangent line trick we get the following:$$ frac{18}{(3-c)(4-c)}-c^2 ge frac{(c+3)}{2}Leftarrow Rightarrow$$ $$ c(c-1)^2(2c-9) le 0$$then the proof claims "and the conclusion follows by summing", now, two things are really confusing to me, first of I don't understand how the "tangent line trick" is applyed here i.e. where does the $frac{c+3}{2}$ comes from...
Apart from that our last inequality, i.e. $c(c-1)^2(2c-9) le 0$ holds only for $0 le c le frac{9}{2}$, so I'm confused since we don't have such bounds on c (nor on a and b), am I not getting this right or is the proof incomplete?
You can find everything I wrote here (page 5): Olympiad Inequalities by Evan Chen
inequality contest-math tangent-line-method
edited Nov 28 '18 at 22:20
Michael Rozenberg
96.3k1588186
asked Nov 27 '18 at 20:52
Spasoje Durovic
33410
add a comment |
I will first state the "trick":
we fix $a=frac{a_1+a_2+...+a_n}{n}, $If $f$ is not convex we can sometimes prove:$$f(x)ge f(a)+f'(a)(x-a) $$ If this manages to hold for all x, then summing up the inequality will give us the desired conclusion.
Then we have the following example:
We want to show, given that $a+b+c=3$: $$sum_{cyc}(frac{18}{(3-c)(4-c)}-c^2) ge 6 $$Using the tangent line trick we get the following:$$ frac{18}{(3-c)(4-c)}-c^2 ge frac{(c+3)}{2}Leftarrow Rightarrow$$ $$ c(c-1)^2(2c-9) le 0$$then the proof claims "and the conclusion follows by summing", now, two things are really confusing to me, first of I don't understand how the "tangent line trick" is applyed here i.e. where does the $frac{c+3}{2}$ comes from...
Apart from that our last inequality, i.e. $c(c-1)^2(2c-9) le 0$ holds only for $0 le c le frac{9}{2}$, so I'm confused since we don't have such bounds on c (nor on a and b), am I not getting this right or is the proof incomplete?
You can find everything I wrote here (page 5): Olympiad Inequalities by Evan Chen
inequality contest-math tangent-line-method
edited Nov 28 '18 at 22:20
Michael Rozenberg
96.3k1588186
asked Nov 27 '18 at 20:52
Spasoje Durovic
33410
add a comment |
I will first state the "trick":
we fix $a=frac{a_1+a_2+...+a_n}{n}, $If $f$ is not convex we can sometimes prove:$$f(x)ge f(a)+f'(a)(x-a) $$ If this manages to hold for all x, then summing up the inequality will give us the desired conclusion.
Then we have the following example:
We want to show, given that $a+b+c=3$: $$sum_{cyc}(frac{18}{(3-c)(4-c)}-c^2) ge 6 $$Using the tangent line trick we get the following:$$ frac{18}{(3-c)(4-c)}-c^2 ge frac{(c+3)}{2}Leftarrow Rightarrow$$ $$ c(c-1)^2(2c-9) le 0$$then the proof claims "and the conclusion follows by summing", now, two things are really confusing to me, first of I don't understand how the "tangent line trick" is applyed here i.e. where does the $frac{c+3}{2}$ comes from...
Apart from that our last inequality, i.e. $c(c-1)^2(2c-9) le 0$ holds only for $0 le c le frac{9}{2}$, so I'm confused since we don't have such bounds on c (nor on a and b), am I not getting this right or is the proof incomplete?
You can find everything I wrote here (page 5): Olympiad Inequalities by Evan Chen
inequality contest-math tangent-line-method
edited Nov 28 '18 at 22:20
Michael Rozenberg
96.3k1588186
asked Nov 27 '18 at 20:52
Spasoje Durovic
33410
I will first state the "trick":
we fix $a=frac{a_1+a_2+...+a_n}{n}, $If $f$ is not convex we can sometimes prove:$$f(x)ge f(a)+f'(a)(x-a) $$ If this manages to hold for all x, then summing up the inequality will give us the desired conclusion.
Then we have the following example:
We want to show, given that $a+b+c=3$: $$sum_{cyc}(frac{18}{(3-c)(4-c)}-c^2) ge 6 $$Using the tangent line trick we get the following:$$ frac{18}{(3-c)(4-c)}-c^2 ge frac{(c+3)}{2}Leftarrow Rightarrow$$ $$ c(c-1)^2(2c-9) le 0$$then the proof claims "and the conclusion follows by summing", now, two things are really confusing to me, first of I don't understand how the "tangent line trick" is applyed here i.e. where does the $frac{c+3}{2}$ comes from...
Apart from that our last inequality, i.e. $c(c-1)^2(2c-9) le 0$ holds only for $0 le c le frac{9}{2}$, so I'm confused since we don't have such bounds on c (nor on a and b), am I not getting this right or is the proof incomplete?
You can find everything I wrote here (page 5): Olympiad Inequalities by Evan Chen
inequality contest-math tangent-line-method
inequality contest-math tangent-line-method
edited Nov 28 '18 at 22:20
Michael Rozenberg
96.3k1588186
asked Nov 27 '18 at 20:52
Spasoje Durovic
33410
edited Nov 28 '18 at 22:20
Michael Rozenberg
96.3k1588186
asked Nov 27 '18 at 20:52
Spasoje Durovic
33410
edited Nov 28 '18 at 22:20
Michael Rozenberg
96.3k1588186
edited Nov 28 '18 at 22:20
Michael Rozenberg
96.3k1588186
edited Nov 28 '18 at 22:20
Michael Rozenberg
96.3k1588186
96.3k1588186
asked Nov 27 '18 at 20:52
Spasoje Durovic
33410
asked Nov 27 '18 at 20:52
Spasoje Durovic
33410
asked Nov 27 '18 at 20:52
Spasoje Durovic
33410
33410
add a comment |
add a comment |
1 Answer
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The inequality, which you want to prove is wrong. Try $crightarrow3^+$.
For positive variables it's true and you wrote the proof.
By the way, you can not use the theorem 2.8 in the contest, otherwise, you need to write a proof of this theorem, which is not so easy. I mean the theorem from this book.
answered Nov 27 '18 at 22:00
Michael Rozenberg
96.3k1588186
add a comment |
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The inequality, which you want to prove is wrong. Try $crightarrow3^+$.
For positive variables it's true and you wrote the proof.
By the way, you can not use the theorem 2.8 in the contest, otherwise, you need to write a proof of this theorem, which is not so easy. I mean the theorem from this book.
answered Nov 27 '18 at 22:00
Michael Rozenberg
96.3k1588186
add a comment |
The inequality, which you want to prove is wrong. Try $crightarrow3^+$.
For positive variables it's true and you wrote the proof.
By the way, you can not use the theorem 2.8 in the contest, otherwise, you need to write a proof of this theorem, which is not so easy. I mean the theorem from this book.
answered Nov 27 '18 at 22:00
Michael Rozenberg
96.3k1588186
add a comment |
The inequality, which you want to prove is wrong. Try $crightarrow3^+$.
For positive variables it's true and you wrote the proof.
By the way, you can not use the theorem 2.8 in the contest, otherwise, you need to write a proof of this theorem, which is not so easy. I mean the theorem from this book.
answered Nov 27 '18 at 22:00
Michael Rozenberg
96.3k1588186
The inequality, which you want to prove is wrong. Try $crightarrow3^+$.
For positive variables it's true and you wrote the proof.
By the way, you can not use the theorem 2.8 in the contest, otherwise, you need to write a proof of this theorem, which is not so easy. I mean the theorem from this book.
answered Nov 27 '18 at 22:00
Michael Rozenberg
96.3k1588186
edited Nov 27 '18 at 22:07
answered Nov 27 '18 at 22:00
Michael Rozenberg
96.3k1588186
answered Nov 27 '18 at 22:00
Michael Rozenberg
96.3k1588186
answered Nov 27 '18 at 22:00
Michael Rozenberg
96.3k1588186
96.3k1588186
add a comment |
add a comment |
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