Proofing an inequality
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Assume we have three points $ a , b , c $ where $a < b < c$ and $b-a=c-b=h $
How can I proof for every $x in [a,b]$ we have $|(x-a)(x-b)(x-c)| leq frac{2sqrt{3}}{9}h^{3}$
linear-algebra
edited Nov 16 at 20:45
mrtaurho
2,5811827
asked Nov 16 at 20:44
Mohamad Abasi
62
add a comment |
up vote
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down vote
favorite
Assume we have three points $ a , b , c $ where $a < b < c$ and $b-a=c-b=h $
How can I proof for every $x in [a,b]$ we have $|(x-a)(x-b)(x-c)| leq frac{2sqrt{3}}{9}h^{3}$
linear-algebra
edited Nov 16 at 20:45
mrtaurho
2,5811827
asked Nov 16 at 20:44
Mohamad Abasi
62
1
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
Nov 16 at 20:46
1
Can you do the case $(a,b,c)=(-1,0,1)$?
– Lord Shark the Unknown
Nov 16 at 20:47
I tried to break problem for $a<x<b$ and $b<x<c$ but the only thing I get its $|(x-a)(x-b)(x-c)| leq 2h^{3}$
– Mohamad Abasi
Nov 16 at 20:55
Have you tried elementary calculus to find max of $(x-a)(x-b)(x-c)$? Note that for $ale xle b$, the expression is always non-negative.
– herb steinberg
Nov 16 at 22:13
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Assume we have three points $ a , b , c $ where $a < b < c$ and $b-a=c-b=h $
How can I proof for every $x in [a,b]$ we have $|(x-a)(x-b)(x-c)| leq frac{2sqrt{3}}{9}h^{3}$
linear-algebra
edited Nov 16 at 20:45
mrtaurho
2,5811827
asked Nov 16 at 20:44
Mohamad Abasi
62
Assume we have three points $ a , b , c $ where $a < b < c$ and $b-a=c-b=h $
How can I proof for every $x in [a,b]$ we have $|(x-a)(x-b)(x-c)| leq frac{2sqrt{3}}{9}h^{3}$
linear-algebra
linear-algebra
edited Nov 16 at 20:45
mrtaurho
2,5811827
asked Nov 16 at 20:44
Mohamad Abasi
62
edited Nov 16 at 20:45
mrtaurho
2,5811827
asked Nov 16 at 20:44
Mohamad Abasi
62
edited Nov 16 at 20:45
mrtaurho
2,5811827
edited Nov 16 at 20:45
mrtaurho
2,5811827
edited Nov 16 at 20:45
mrtaurho
2,5811827
2,5811827
asked Nov 16 at 20:44
Mohamad Abasi
62
asked Nov 16 at 20:44
Mohamad Abasi
62
asked Nov 16 at 20:44
Mohamad Abasi
62
62
1
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
Nov 16 at 20:46
1
Can you do the case $(a,b,c)=(-1,0,1)$?
– Lord Shark the Unknown
Nov 16 at 20:47
I tried to break problem for $a<x<b$ and $b<x<c$ but the only thing I get its $|(x-a)(x-b)(x-c)| leq 2h^{3}$
– Mohamad Abasi
Nov 16 at 20:55
Have you tried elementary calculus to find max of $(x-a)(x-b)(x-c)$? Note that for $ale xle b$, the expression is always non-negative.
– herb steinberg
Nov 16 at 22:13
add a comment |
1
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
Nov 16 at 20:46
1
Can you do the case $(a,b,c)=(-1,0,1)$?
– Lord Shark the Unknown
Nov 16 at 20:47
I tried to break problem for $a<x<b$ and $b<x<c$ but the only thing I get its $|(x-a)(x-b)(x-c)| leq 2h^{3}$
– Mohamad Abasi
Nov 16 at 20:55
Have you tried elementary calculus to find max of $(x-a)(x-b)(x-c)$? Note that for $ale xle b$, the expression is always non-negative.
– herb steinberg
Nov 16 at 22:13
1
1
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
Nov 16 at 20:46
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
Nov 16 at 20:46
1
1
Can you do the case $(a,b,c)=(-1,0,1)$?
– Lord Shark the Unknown
Nov 16 at 20:47
Can you do the case $(a,b,c)=(-1,0,1)$?
– Lord Shark the Unknown
Nov 16 at 20:47
I tried to break problem for $a<x<b$ and $b<x<c$ but the only thing I get its $|(x-a)(x-b)(x-c)| leq 2h^{3}$
– Mohamad Abasi
Nov 16 at 20:55
I tried to break problem for $a<x<b$ and $b<x<c$ but the only thing I get its $|(x-a)(x-b)(x-c)| leq 2h^{3}$
– Mohamad Abasi
Nov 16 at 20:55
Have you tried elementary calculus to find max of $(x-a)(x-b)(x-c)$? Note that for $ale xle b$, the expression is always non-negative.
– herb steinberg
Nov 16 at 22:13
Have you tried elementary calculus to find max of $(x-a)(x-b)(x-c)$? Note that for $ale xle b$, the expression is always non-negative.
– herb steinberg
Nov 16 at 22:13
add a comment |
1 Answer
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$$f(x)=|(x-a)(x-b)(x-c)|=(x-a)(x-b)(x-c)=(x-a)(b-x)(c-x)tag{1}$$
Introduce substitution:
$$b-x=u, quad uin[0,h]tag{2}$$
$$x-a=(b-a)-(b-x)=h-utag{3}$$
$$c-x=(c-b)+(b-x)=h+utag{4}$$
Now replace (2,3,4) into (1):
$$f(u)=(h-u)u(h+u)=u(h^2-u^2)=uh^2-u^3tag{5}$$
$$f'(u)=h^2-3u^2=0implies u=frac{hsqrt{3}}{3}tag{6}$$
$$f(u)|_{u=frac{hsqrt{3}}{3}}=frac{2sqrt{3}}{9}h^{3}tag{7}$$
Obviously $f''(u)=-3u^2<0$ for all values of $une0$. This proves that (7) is indeed the maximum value.
answered Nov 17 at 18:29
Oldboy
5,7331628
add a comment |
1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
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up vote
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$$f(x)=|(x-a)(x-b)(x-c)|=(x-a)(x-b)(x-c)=(x-a)(b-x)(c-x)tag{1}$$
Introduce substitution:
$$b-x=u, quad uin[0,h]tag{2}$$
$$x-a=(b-a)-(b-x)=h-utag{3}$$
$$c-x=(c-b)+(b-x)=h+utag{4}$$
Now replace (2,3,4) into (1):
$$f(u)=(h-u)u(h+u)=u(h^2-u^2)=uh^2-u^3tag{5}$$
$$f'(u)=h^2-3u^2=0implies u=frac{hsqrt{3}}{3}tag{6}$$
$$f(u)|_{u=frac{hsqrt{3}}{3}}=frac{2sqrt{3}}{9}h^{3}tag{7}$$
Obviously $f''(u)=-3u^2<0$ for all values of $une0$. This proves that (7) is indeed the maximum value.
answered Nov 17 at 18:29
Oldboy
5,7331628
add a comment |
up vote
0
down vote
$$f(x)=|(x-a)(x-b)(x-c)|=(x-a)(x-b)(x-c)=(x-a)(b-x)(c-x)tag{1}$$
Introduce substitution:
$$b-x=u, quad uin[0,h]tag{2}$$
$$x-a=(b-a)-(b-x)=h-utag{3}$$
$$c-x=(c-b)+(b-x)=h+utag{4}$$
Now replace (2,3,4) into (1):
$$f(u)=(h-u)u(h+u)=u(h^2-u^2)=uh^2-u^3tag{5}$$
$$f'(u)=h^2-3u^2=0implies u=frac{hsqrt{3}}{3}tag{6}$$
$$f(u)|_{u=frac{hsqrt{3}}{3}}=frac{2sqrt{3}}{9}h^{3}tag{7}$$
Obviously $f''(u)=-3u^2<0$ for all values of $une0$. This proves that (7) is indeed the maximum value.
answered Nov 17 at 18:29
Oldboy
5,7331628
add a comment |
up vote
0
down vote
up vote
0
down vote
$$f(x)=|(x-a)(x-b)(x-c)|=(x-a)(x-b)(x-c)=(x-a)(b-x)(c-x)tag{1}$$
Introduce substitution:
$$b-x=u, quad uin[0,h]tag{2}$$
$$x-a=(b-a)-(b-x)=h-utag{3}$$
$$c-x=(c-b)+(b-x)=h+utag{4}$$
Now replace (2,3,4) into (1):
$$f(u)=(h-u)u(h+u)=u(h^2-u^2)=uh^2-u^3tag{5}$$
$$f'(u)=h^2-3u^2=0implies u=frac{hsqrt{3}}{3}tag{6}$$
$$f(u)|_{u=frac{hsqrt{3}}{3}}=frac{2sqrt{3}}{9}h^{3}tag{7}$$
Obviously $f''(u)=-3u^2<0$ for all values of $une0$. This proves that (7) is indeed the maximum value.
answered Nov 17 at 18:29
Oldboy
5,7331628
$$f(x)=|(x-a)(x-b)(x-c)|=(x-a)(x-b)(x-c)=(x-a)(b-x)(c-x)tag{1}$$
Introduce substitution:
$$b-x=u, quad uin[0,h]tag{2}$$
$$x-a=(b-a)-(b-x)=h-utag{3}$$
$$c-x=(c-b)+(b-x)=h+utag{4}$$
Now replace (2,3,4) into (1):
$$f(u)=(h-u)u(h+u)=u(h^2-u^2)=uh^2-u^3tag{5}$$
$$f'(u)=h^2-3u^2=0implies u=frac{hsqrt{3}}{3}tag{6}$$
$$f(u)|_{u=frac{hsqrt{3}}{3}}=frac{2sqrt{3}}{9}h^{3}tag{7}$$
Obviously $f''(u)=-3u^2<0$ for all values of $une0$. This proves that (7) is indeed the maximum value.
answered Nov 17 at 18:29
Oldboy
5,7331628
answered Nov 17 at 18:29
Oldboy
5,7331628
answered Nov 17 at 18:29
Oldboy
5,7331628
answered Nov 17 at 18:29
Oldboy
5,7331628
5,7331628
add a comment |
add a comment |
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Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
Nov 16 at 20:46
1
Can you do the case $(a,b,c)=(-1,0,1)$?
– Lord Shark the Unknown
Nov 16 at 20:47
I tried to break problem for $a<x<b$ and $b<x<c$ but the only thing I get its $|(x-a)(x-b)(x-c)| leq 2h^{3}$
– Mohamad Abasi
Nov 16 at 20:55
Have you tried elementary calculus to find max of $(x-a)(x-b)(x-c)$? Note that for $ale xle b$, the expression is always non-negative.
– herb steinberg
Nov 16 at 22:13