✔ 最佳答案
1) Accordingly, we know that ‘a>b>c>0’,then let a=3, b=2 and c=1, 3>2>1>0.Compare(a^a)*(b^b)*(c^c) and (a^b)*(b^c)*(c^a). Sub.a=3, b=2 and c= 1 into both algebraic expressions, we have, LHS=(3^3)*(2^2)*(1^1) =108 RHS=(3^2)*(2^1)*(1^3) =18 ∵108>18 ∴ (a^a)*(b^b)*(c^c) >(a^b)*(b^c)*(c^a) 2) Accordingly, weknow that ‘a+b=c’, then let a=1, b=2 and c=3, 1+2=3.Compare a^(2/3)+b^(2/3) and c^(2/3). Sub. a=1, b=2 and c=3 into both algebraic expressions,we have, LHS=1^(2/3)+2^(2/3) =2.59 (corr.to 3 sig. fig.) RHS=3^(2/3) =2.08(corr. to 3 sig. fig.) ∵2.59>2.08
∴ a^(2/3)+b^(2/3) >c^(2/3)
2014-02-09 23:13:34 補充:
2) Also,consider when a,b,c are all equal to 0,and 0+0=0.
Compare a^(2/3)+b^(2/3) and c^(2/3)
Sub. a=0,b=0, and c=0 into both sides.
LHS=0^(2/3)+0^(2/3)
=0
RHS=^(2/3)
=0
∵0=0
∴ a^(2/3)+b^(2/3) =c^(2/3)
In conclusion, a^(2/3)+b^(2/3) ≥ c^(2/3)