inequalities questions.. (1 Viewer)

[haboozin]

just a couple of questions:


1.If a>0 and b>0 and c>0
show

a/b + b/c + c/a >= 3



2. If a, b are >0 and a+b = t show that

1/a² + 1/b² >= 8/t²



3.
show that e^x > 1 + x

can i do this one graphically?
can we solve inequalities graphically?

 

[Slidey]

booz: Sorry. I'll let somebody who's going to give you their full attention do the problems. You should still be able to do it via LHS-RHS though.

 

[haboozin]

LHS-RHS=a^-2+b^-2-8/(a+b)

LHS-RHS=a^-2+b^-2-8/(a+b)²

 

[tiggerfamilytre]

hopefully this helps, i think graphical methods are valid (you probably had the same answer as me in 3)

 

[haboozin]

hopefully this helps, i think graphical methods are valid (you probably had the same answer as me in 3)

i dont get the first 1?

 

[KFunk]

He's using the general form of the cauchy inequality:

(a₁ + a₂ + ... + a_n)/n ≥ n-root(a₁a₂...a_n)

so (a₁ + a₂ + ... + a_n) ≥ n[n-root(a₁a₂...a_n)]

 

[tiggerfamilytre]

it's just using the result (x + y + z)/3 >= (xyz)^[1/3] substituting a/b, b/c, c/a for x, y, z, so the RHS cancels to become 1, and you just multiply the 3 across.

 

[turtle_2468]

He's using the general form of the cauchy inequality:

(a₁ + a₂ + ... + a_n)/n ≥ n-root(a₁a₂...a_n)

so (a₁ + a₂ + ... + a_n) ≥ n[n-root(a₁a₂...a_n)]

you mean the specific form of the cauchy inequality, otherwise known as the AM-GM?

 

[haboozin]

you mean the specific form of the cauchy inequality, otherwise known as the AM-GM?

i know why i couldnt do it now because of the a + b + c >=3 cubedroot(abc)


but is cauchy in the syllabus?

 

[turtle_2468]

i know why i couldnt do it now because of the a + b + c >=3 cubedroot(abc)


but is cauchy in the syllabus?

as far as I know, no.

 

[haboozin]

can anyone do this non graphically:

x >= 3sin@/(2 + cos@)

 

[richz]

let f(x) = 3sin@/(2+cos@) -x
then derive and find if its a max or min etc etc

 

[Sepulchres]

as far as I know, no.

However, you should know it as it helps to solve quite a few inequality questions.

 

[turtle_2468]

however, you probably can't assume cauchy anyway. Because it's nowhere near the reach of the syllabus you'd probably have to prove it (2 pages+) in order to use it..
The AM-GM is useful, and probably the only thing (short of them guiding you through something) that you'll need for 4U.

 

[justchillin]

He's using the general form of the cauchy inequality:

(a₁ + a₂ + ... + a_n)/n ≥ n-root(a₁a₂...a_n)

so (a₁ + a₂ + ... + a_n) ≥ n[n-root(a₁a₂...a_n)]

With the original quesiton I wouldnt advise just quoting the cauchy inequality... Id prove the a+b+c>3rt(abc) then sub a=whatever u need... etc

I say this becasue I remember seeing this very question in a trial paper somewhere and the solns explicity said that u cant just quote something like that...
In saying this, it is most likely the quesiton wil have a lead-in...

 

[haboozin]

if they give you:

prove:

x^3 + y^3 + z^3 >= 3xyz


without any lead in (i know that is pretty farfetched.

could we do some dodgy factorizing


ie:

x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 -xy - xz - zy)

that would make the job alot easier.....

 

[who_loves_maths]

Originally Posted by haboozin
2. If a, b are >0 and a+b = t show that

1/a^2 + 1/b^2 >= 8/t^2

Originally Posted by tiggerfamilytre
hopefully this helps, i think graphical methods are valid (you probably had the same answer as me in 3)
...

^ tigger, your solution to Question 2 {from Terry Lee} is incorrect.


this is what it's supposed to have said:

LHS = 1/a^2 + 1/b^2 > 2/(ab) .................................................. [AM-GM inequality]

but since (a + b)^2/4 > ab, then, 1/(ab) > 4/(a + b)^2
-----> ie. 2/(ab) > 8/(a + b)^2 = RHS

Therefore: 1/a^2 + 1/b^2 > 8/(a + b)^2 = 8/t^2

 

[香港!]

wow who_loves_maths!!
ur back!!
where hav u been???
studying i guess?

 

[香港!]

^ lol, yea i have. well... sort of...

^
must've been some intense studying @_@
its been nearly 2 months lol
how do u do that? without coming online at all??
i wish i could be as focused as u

 

[who_loves_maths]

^ lol, yea i have. well... sort of...