Homework Help

Let a and b be positive real numbers such that                          ...

user profile pic

mathsiq | Student | eNoter

Posted June 23, 2013 at 2:18 AM via web

dislike 0 like

Let a and b be positive real numbers such that

                              a^2 + b^2 =1

Prove that 

                              a^4 + b^4 `>= 1/2`

Please help me with this question...thankyou

Tagged with indices, inequations, math, maths, powers

3 Answers | Add Yours

Top Answer

user profile pic

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted June 23, 2013 at 2:41 AM (Answer #1)

dislike 1 like

`a^2+b^2 = 1`

We know that `a^2+2ab+b^2 = (a+b)^2`

Also `(a+b)^2>=0`

`(a^2+b^2+2ab) >=0`

`1+2ab >=0`

`2ab >=-1`

`4a^2b^2>=1`

 

`a^4+b^4 = (a^2+b^2)^2-2a^2b^2`

 

`a^4+b^4 = 1-2a^2b^2`

We found that `4a^2b^2>=1` .

So we can say `2a^2b^2>=1/2`

 

`a^4+b^4 = 1-2a^2b^2`

`a^4+b^4 >=1-1/2`

`a^4+b^4 >=1/2`

 

So the answer is proved.

3 Replies | Hide Replies

user profile pic

degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted June 23, 2013 at 3:10 AM (Reply #1)

dislike 0 like

This doesn't work. `2ab>=-1` doesn't imply that

`4a^2b^2>=1.` To see this, let `a=1,b=0.` Then

`2ab=0>=-1,` but `4a^2b^2=0<1.` I think the inequality

we need is actually `2a^2b^2<=1/2,` since below if we subtract

something less than `1/2` from 1, we end up with something

greater than `1/2,` which is what we want. If `2a^2b^2>=1/2,`

then we would end up with `a^4+b^4<=1/2,` which is wrong.

user profile pic

mathsiq | Student | eNoter

Posted June 23, 2013 at 10:11 AM (Reply #2)

dislike 0 like

Thank you for this response. It really helped.

user profile pic

mathsiq | Student | eNoter

Posted June 23, 2013 at 10:24 AM (Reply #3)

dislike 0 like

and i wrote that a and b were positive

user profile pic

degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted June 23, 2013 at 2:55 AM (Answer #2)

dislike 1 like

If you know trigonometry there's a nice solution. Since `a^2+b^2=1,` then we must have `a=sin theta` and `b=cos theta` for some `0<=theta<=2pi.` Then using the identity `1/2sin(2theta)=sinthetacostheta` and the fact that `sin^2(2theta)<=1` for all `theta,` we see that

`a^4+b^4=1-2(ab)^2=1-2(sintheta cos theta)^2`

`=1-2(1/2 sin(2theta))^2`

`=1-1/2 sin^2(2theta)`

`>=1-1/2`

`=1/2.`

Sources:

user profile pic

aruv | High School Teacher | Valedictorian

Posted June 23, 2013 at 5:38 AM (Answer #3)

dislike 0 like

a,b are positive real numbers i.e. `a,b>=0`

`(a-b)^2>=0 `

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

`a^2+b^2>=2ab`

`1>=2ab`

`(1/2)>=ab`

`(1/2)xx(1/2)>=(1/2)(ab)>=(ab)(ab)=a^2b^2`

`(1/4)>=a^2b^2`                              (i)

using following an identity 

`a^4+b^4=(a^2+b^2)^2-2a^2b^2`

we can prove that

`a^4+b^4=1-2a^2b^2`          (ii)

From (i)

`(1/4)>=a^2b^2`

`-(1/4)<=-a^2b^2`

`-2xx(1/4)<=-2a^2b^2`

`1-2/4<=1-2a^2b^2`

`1-2a^2b^2>=1/2`                                   (iii)

using (iii) in (ii)

`a^4+b^4=1-2a^2b^2>=1/2`

`a^4+b^4>=1/2`

what you wish to prove.

 

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes