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

### 3 Answers | Add Yours

`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 ▲

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.

Thank you for this response. It really helped.

and i wrote that a and b were positive

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:**

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.

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