341Stronger: if x+y+z>0 then x3+y3+z3≥3xyz
A very simple one... specially the 2nd one...
a) if x+y+z=0, x3+y3+z3=3xyz
b) Prove that if x, y, z are positive,
x3+y3+z3≥3xyz
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10 Answers
1that is why I asked the question straightaway
instead of applying it [3]
21a)
x+y+z=0
x+y=-z
(x+y)3 = (-z)3
x3+y3+3xy(x+y) = -z3
x3+y3+3xy(-z) = -z3
x3+ y3 + z3 = 3xyz
21b)
x+y+z≥0
x+y≥-z
(x+y)3 ≥ (-z)3
x3+y3+3xy(x+y) ≥ -z3
x3+y3+3xy(-z) ≥ -z3
x3+ y3 + z3 ≥ 3xyz
1(b)
AM ≥ GM
x3+y3+z33 ≥ (x3y3z3)1/3
→ x3+y3+z3 ≥ 3xyz
39Ahh..the way sibal solved it sparked old memories of class 9..lol.
A little addition to the b) part which sibal solved.
We can't substitute x + y = -z can we? As it is an inequality...
x3 + y3 + 3xy(x + y) ≥ -z3
=> x3 + y3 + z3 ≥ -3xy(x + y)
Note from the inequality :
x + y ≥ -z
=>3(x + y) ≥ -3z
=> 3xy(x + y) ≥ -3xyz
=> -3xy(x + y) ≤ 3xyz
So x3 + y3 + z3 ≥ -3xy(x + y)
How would we proceed from here?
Another thing Tushar brought to my notice...
If x, y and z are positive, how in the world can they ever be zero???
x + y + z > 0 hoga na?