proofs 4U (1 Viewer)
- Thread starter kractus
- Start date
Would the following working help with Question 13 (d)?
From part (i):
^2\le a^2)
)
Similarly:
)
)
Adding (1), (2) and (3) together:
)
^2=a^2+b^2+c^2+2ab+2bc+2ca\:\:\:\left(5\right))
Substituting (4) into (5):
^2\le 2ab+2bc+2ca+2ab+2bc+2ca)
^2\le 4\left(ab+bc+ca\right))
From part (i):
Similarly:
Adding (1), (2) and (3) together:
Substituting (4) into (5):
Masaken
Unknown Member
funny enough, i did this as exact question as part of my school hw a few days ago
14ci - move the RHS onto the LHS, so you can have RTP a^3 + b^3 - (a/c + b/c)abc > 0 (pretend = sign is under the >)
expand the brackets out, then it's a matter of factorising through grouping in pairs, then you can prove the inequality for > 0, then you can do similar for the other two expressions the question asks you to do
14cii - use part i (bc of 'hence'). in part i, you have one inequality wiht a^3 + b^3, and you made two more for b^3 + c^3 and c^3 + a^3. add those three together, then from there you will get the inequality
