Math 4U Proof, Inequalities

[Dean54321]

Part i) 1/a + 1/b >= 4/(a+b)
Part ii) 1/a^2 + 1/b^2 >= 8/(a+b)^2

I have completed part i) but part ii) is confusing me. My main attempt was as such:
1/a + 1/b >= 4/(a+b), using i
1/a^2 + 2/ab + 1/b^2 >= 16/(a+b)^2
1/a^2 + 1/b^2 >= 16/(a+b)^2 - 2/ab

R.T.P 16/(a+b)^2 - 2/ab >= 8/(a+b)^2
LHS - RHS = 8/(a+b)^2 -2/ab
= 4ab-2a^2-2b^2/(a+b)^2ab

but the top, by AM/GM inequality indicates that LHS - RHS <= 0, and so I gets stuck here.

 

[topsquark]

Part i) 1/a + 1/b >= 4/(a+b)
Part ii) 1/a^2 + 1/b^2 >= 8/(a+b)^2

I have completed part i) but part ii) is confusing me. My main attempt was as such:
1/a + 1/b >= 4/(a+b), using i
1/a^2 + 2/ab + 1/b^2 >= 16/(a+b)^2
1/a^2 + 1/b^2 >= 16/(a+b)^2 - 2/ab

R.T.P 16/(a+b)^2 - 2/ab >= 8/(a+b)^2
LHS - RHS = 8/(a+b)^2 -2/ab
= 4ab-2a^2-2b^2/(a+b)^2ab

but the top, by AM/GM inequality indicates that LHS - RHS <= 0, and so I gets stuck here.

Some advice: You will find these to be much easier to add the fractions on the LHS:
[imath]\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{a + b}{ab}[/imath]

Then
[imath]\dfrac{a + b}{ab} \geq \dfrac{4}{a + b}[/imath]

Now use AM-GM.

-Dan

 

[Steven G]

If two quantities, say b and c, are both less than a, then you can not conclude that b<c or c<b. In fact they may both be equal to one another.

Also 1/ab\(\displaystyle \ne \dfrac {1}{ab}\). You need to write 1/(ab)

 

[Steven G]

Another problem with your work.

If a <b, it does NOT follow that a^2<b^2.

Clearly -7 < 2, but 49 <4 is not true