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Hy ,
(a;b;c) ≥ 0
Prove That :
i) 4(a³+b^3)≥ (a+b)³
ii) 9(a³+b³+c³)≥ (a+b+c)³
iii) (a+b+c)³ ≥ a³+b³+c³+24abc
Hi @ssy;
For i )
Expand the RHS:
3 >0 and (a-b)^2 > 0 and (a+b)>0 So 3(a-b)^2(a+b)>0, and we are done!
For iii)
Expand the LHS:
Clean up a little:
Subtract 6abc from both sides:
Divide by 3:
We can prove A with the AM-GM.
So A is proved and so is iii, we are done.
Now for ii ...
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi bobbym ;
For ii)
L'inégo is equivalent : 8(a^3+b^3+c^3) ≥ 3∑ab(a+b)+6abc
by i) and AM-GM : 2(a^3+b^3+c^3) ≥ 6abc , and we are done!