Need help plz (1 Viewer)

[Bellow]

Further questions 8
1.Show that (lx+my+nz)²≤(l²+m²+n²)(x²+y²+z²)
Deduce that
(a) (a+b+c)²≤3(a²+b²+c²)
(b) (a³+b³+c³)≤(a²+b²+c²)(a^4+b^4+c^4)

2. If a>0, b>0, c>0 and d>0, show that

16/(a+b+c+d)≤3/(b+c+d) + 3/(a+c+d) + 3/(a+b+d) + 3/(a+b+c)≤1/a + 1/b +1/c + 1/d

if any1 can post solutions up thx very much

(btw, does any1 kno which book these questions r from?)

 

[SeDaTeD]

Well, not really a 4u method, but for the first bit, let a be the vector li + mj + nk, b = xi + yj + zk.
(lx+my+nz)² = (a.b)^2 = (|a||b|cos(theta))^2 <= (|a||b|)^2 = |a|^2|b|^2 = (l²+m²+n²)(x²+y²+z²) . Theta being the angle between a and b.
I forgot those 4u methods, ohwell.

 

[haboozin]

Further questions 8
1.Show that (lx+my+nz)²≤(l²+m²+n²)(x²+y²+z²)
Deduce that
(a) (a+b+c)²≤3(a²+b²+c²)
(b) (a³+b³+c³)≤(a²+b²+c²)(a^4+b^4+c^4)

2. If a>0, b>0, c>0 and d>0, show that

16/(a+b+c+d)≤3/(b+c+d) + 3/(a+c+d) + 3/(a+b+d) + 3/(a+b+c)≤1/a + 1/b +1/c + 1/d

if any1 can post solutions up thx very much

(btw, does any1 kno which book these questions r from?)

1. expand, factorize

you get 2xmy + 2lxnz + 2mynz <= y^2l^2 + z^2l^2 + m^2x^2 + m^2z^2 + n^2x^2 + n^2y^

what do you see? whole lots of factorization

0<= (yl -mx)^2 + (zl - nx)^2 + (mz - ny)^2

since l, m, n, x, y, z are positive numbers and are real (has to be or otherwize its not in 4unit) this is true


(a) (a+b+c)²≤3(a²+b²+c²)



a^2 + b^2 + c^2 + 2ab + 2bc + 2ac <= 3a^2 + 3b^2 + 3c^2
ab + ac + bc <=a^2 + b^2 + c^2

now a^2 + b^2 >= 2ab
b^2 + c^2 >= 2bc
c^2 + a^2 >= 2ac

2(a^2 + b^2 + c^2)>= 2(ab + bc + ac)

so a^2 + b^2 + c^2 >=ab + bc + ac




you have "further questions" and chapter 8 ... so Arnold Cambridge?

 

[KFunk]

ok i think u have to slow down a bit.....im a lil lost sry.....i dont get the first 1 and
i dont get y
a^2 + b^2 + c^2 + 2ab + 2bc + 2ac <= 3a^2 + 3b^2 + 3c^2

That's just an expansion of (a + b + c)² &le; 3(a² + b² + c²)

On the left side you have:
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

On the right side you have:
3(a² + b² + c²) = 3a² + 3b² + 3c²

voila


(edit: The servers are retarded)

 

[Bellow]

1. expand, factorize

you get 2xmy + 2lxnz + 2mynz <= y^2l^2 + z^2l^2 + m^2x^2 + m^2z^2 + n^2x^2 + n^2y^

what do you see? whole lots of factorization

0<= (yl -mx)^2 + (zl - nx)^2 + (mz - ny)^2

since l, m, n, x, y, z are positive numbers and are real (has to be or otherwize its not in 4unit) this is true


(a) (a+b+c)²≤3(a²+b²+c²)



a^2 + b^2 + c^2 + 2ab + 2bc + 2ac <= 3a^2 + 3b^2 + 3c^2
ab + ac + bc <=a^2 + b^2 + c^2

now a^2 + b^2 >= 2ab
b^2 + c^2 >= 2bc
c^2 + a^2 >= 2ac

2(a^2 + b^2 + c^2)>= 2(ab + bc + ac)

so a^2 + b^2 + c^2 >=ab + bc + ac




you have "further questions" and chapter 8 ... so Arnold Cambridge?

ok i think u have to slow down a bit.....im a lil lost sry.....i dont get the first 1 and
i dont get y
a^2 + b^2 + c^2 + 2ab + 2bc + 2ac <= 3a^2 + 3b^2 + 3c^2

 

[Bellow]

Well, not really a 4u method, but for the first bit, let a be the vector li + mj + nk, b = xi + yj + zk.
(lx+my+nz)² = (a.b)^2 = (|a||b|cos(theta))^2 <= (|a||b|)^2 = |a|^2|b|^2 = (l²+m²+n²)(x²+y²+z²) . Theta being the angle between a and b.
I forgot those 4u methods, ohwell.

sry i dun think thats a 4u method of inequalities

 

[Bellow]

That's just an expansion of (a + b + c)² &le; 3(a² + b² + c²)

On the left side you have:
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

On the right side you have:
3(a² + b² + c²) = 3a² + 3b² + 3c²

voila


(edit: The servers are retarded)

wait.....did he end up proving....
(a+b+c)²≤3(a²+b²+c²)?

 

[Templar]

Well, not really a 4u method, but for the first bit, let a be the vector li + mj + nk, b = xi + yj + zk.
(lx+my+nz)² = (a.b)^2 = (|a||b|cos(theta))^2 <= (|a||b|)^2 = |a|^2|b|^2 = (l²+m²+n²)(x²+y²+z²) . Theta being the angle between a and b.
I forgot those 4u methods, ohwell.

You could have just quoted Cauchy instead, and probably get it out in less lines.

 

[haboozin]

wait.....did he end up proving....
(a+b+c)²≤3(a²+b²+c²)?

yea

consider this:

a^2 + b^2 >= 2ab

a^2 -2ab + b^2 >= 0
...

(a-b)^2 >=0

true since anything squared is >=0

now

concider these three

(1) a^2 + b^2 >= 2ab
(2) b^2 + c^2 >= 2bc
(3) c^2 + a^2 >= 2ac

now add 1, 2, 3 together

2a^2 + 2b^2 + 2c^2 >= 2ab + 2ac + 2bc

now add a^2 + b^2 + c^2 to both sides

3a^2 + 3b^2 + 3c^2 >= a^2 + b^2 + c^2 + 2ab + 2bc + 2ac

factorize both sides

3(a^2 + b^2 + c^2) >= (a + b + c)^2

\

 

[Bellow]

yea

consider this:

a^2 + b^2 >= 2ab

a^2 -2ab + b^2 >= 0
...

(a-b)^2 >=0

true since anything squared is >=0

now

concider these three

(1) a^2 + b^2 >= 2ab
(2) b^2 + c^2 >= 2bc
(3) c^2 + a^2 >= 2ac

now add 1, 2, 3 together

2a^2 + 2b^2 + 2c^2 >= 2ab + 2ac + 2bc

now add a^2 + b^2 + c^2 to both sides

3a^2 + 3b^2 + 3c^2 >= a^2 + b^2 + c^2 + 2ab + 2bc + 2ac

factorize both sides

3(a^2 + b^2 + c^2) >= (a + b + c)^2

\

cool thx man

 

[haboozin]

cool thx man

the question said deduce... so there has to be an easier way to do this using the question before it...


ie .... let lx = a etc or something

but i cant see it, sorry.

 

[Templar]

(a+b+c)^2=(1a+1b+1c)^2
<=(1+1+1)(a^2+b^2+c^2)

 

[haboozin]

(a+b+c)^2=(1a+1b+1c)^2
>=(1+1+1)(a^2+b^2+c^2)

omg so easy...

so stupid of me..

bellow listen to Templar not me