Harder 3U Question (???) (1 Viewer)

[cutemouse]

Could someone please help me with these? Thanks

Suppose that a>0, b>0, c>0

(a) Prove that a^2+b^2 >= 2ab
(b) Hence, or otherwise prove that a^2+b^2+c^2 >=ab+bc+ca
(c) Given a^3+b^3+c^2=(a+b+c)(a^2+b^2+c^2-ab-bc-ca), prove that a^3+b^3+c^3>=3abc

Thanks

 

[Dumbledore]

i haven't learnt harder 3u yet but i think question 1 goes like

(a-b)^2=a^2-2ab+b^2
so a^2+b^2=(a+b)^2+2ab.. since a>0 and b>0 : (a+b)^2>0
hence a^2+b^2 >= 2ab

 

[Dumbledore]

b) from 1: a^2+b^2>=2ab, similarly b^2+c^2>=2cb and c^2+a^2>=2ac

add equations 1,2 and 3 : 2(a^2+b^2+c^2) >= 2ab+2bc+2ac
a^2+b^2+c^2>=ab+bc+ac

 

[Drongoski]

i haven't learnt harder 3u yet but i think question 1 goes like

(a-b)^2=a^2-2ab+b^2
so a^2+b^2=(a+b)^2+2ab.. since a>0 and b>0 : (a+b)^2>0
hence a^2+b^2 >= 2ab

For real a & b, (a-b)² >= 0

i.e a² + b² - 2ab >= 0

.: a² + b² >= 2ab

Here a>0 and b>0 are not a requirement.

 

[Dumbledore]

dam that was stupid of me lol, if i took out a>0 and b>0 it would be right though

 

[Drongoski]

dam that was stupid of me lol, if i took out a>0 and b>0 it would be right though

dw mate. There are cases where a>0, b>0 etc may be important but not this one. I thot bringing attn to this point might benefit viewers.

 

[Trebla]

Could someone please help me with these? Thanks

Suppose that a>0, b>0, c>0

(a) Prove that a^2+b^2 >= 2ab
(b) Hence, or otherwise prove that a^2+b^2+c^2 >=ab+bc+ca
(c) Given a^3+b^3+c^2=(a+b+c)(a^2+b^2+c^2-ab-bc-ca), prove that a^3+b^3+c^3>=3abc

Thanks

(c) I don't know about the given equation, it looks like there's something wrong about it in the LHS.
Note that:
(a + b + c)(a² + b² + c² - ab - ac - bc)
= a³ + ab² + ac² - a²b - a²c - abc + a²b + b³ + bc² - ab² - abc - b²c + a²c + b²c + c³ - abc - ac² - bc²
= a³ - abc + b³ - abc + c³ - abc
= a³ + b³ + c³ - 3abc
.: (a + b + c)(a² + b² + c² - ab - ac - bc) = a³ + b³ + c³ - 3abc
but a² + b² + c² ≥ ab + ac + bc from (b)
=> a² + b² + c² - ab - ac - bc ≥ 0
Also, since a, b, c > 0 then: a + b + c > 0
Hence:
(a + b + c)(a² + b² + c² - ab - ac - bc) ≥ 0
=> a³ + b³ + c³ - 3abc ≥ 0
.: a³ + b³ + c³ ≥ 3abc