21if ai ≥0...i=1,2...n
then a1m+a2m+a3m........anmn≤(a1+a2+a3.........ann)m
if 0<m<1
here a1=3+31/3
a2=3-31/3
so
(3+31/3)1/3 + (3-31/3)1/32≤( 3+31/3+3-31/3 2)1/3
(3+31/3)1/3 + (3-31/3≤2(6/2)1/3
≤2. 31/3
9 Answers
21
341a^3+b^3+c^3-3abc <0 \Leftrightarrow a+b+c<0
With a = \sqrt [3] {3+\sqrt [3] {3}}; b = \sqrt [3] {3-\sqrt [3] {3}}; c= 2 \sqrt [3] 3 it suffices to prove that
3> \sqrt [3] {3(9-\sqrt [3] 9)}
or9>9-\sqrt [3] 9 which is obviously true
1i appreciate the method used by eragon but there is a reason i gave it under calculus ,
can anybody use calculus and solve it for me???
341a^3 + b^3 +c^3-3abc = \frac{1}{2} (a+b+c) [(a-b)^2+(b-c)^2+(c-a)^2] = P(a+b+c)
where P is always positive.
1i'll post the solution of the problem using calculus
consider the function f(x)=3√x
we can easily find (using calculus) that it is concave in the interval (0,+∞), then consider two distinct numbers a,b >0 we have
{f(a)+f(b)}/2 < f[(a+b)/2]
taking a= 3-3√3
and b=3+3√3
we obtain the desired result
this is the solution which was on my mind when i posted this question ,different kind of solutions are still welcome!!!