Inequality.......幫幫手解答.....

1)
Let S1 = a1 + b1 , S2 = a2 + b2 , ... , Sn = an + bn .By AM ≥ GM :(a1 / S1 + a2 / S2 + ... + an / Sn) / n ≥ ⁿ√( a1 a2 ... an / (S1 S2 ... Sn) )
and
(b1 / S1 + b2 / S2 + ... + bn / Sn) / n ≥ ⁿ√( b1 b2 ... bn / (S1 S2 ... Sn) )then
(a1 / S1 + a2 / S2 + ... + an / Sn) / n + (b1 / S1 + b2 / S2 + ... + bn / Sn) / n

≥ ⁿ√( a1 a2 ... an / (S1 S2 ... Sn) ) + ⁿ√( b1 b2 ... bn / (S1 S2 ... Sn) )==>

1 ≥ ( ⁿ√(a1 a2 ... an) + ⁿ√(b1 b2 ... bn) ) / ⁿ√(S1 S2 ... Sn)
==>
ⁿ√( (a1 + b1) (a2 + b2) ... (an + bn) ) ≥ ⁿ√(a1 a2 ... an) + ⁿ√(b1 b2 ... bn)

2)
ⁿ⁺¹√ ( (1 + 1/(n+1))ⁿ⁺² ) = ⁿ⁺¹√ ( (1 + 1/(n+1))ⁿ * (1 + 1/(n+1))² )

By AM ≥ GM ( '=' is not hold here) :

< ( n (1 + 1/(n+1)) + (1 + 1/(n+1))² ) / (n+1)
= ( n + n/(n+1) + 1 + 2/(n+1) + 1/(n+1)² ) / (n+1)
= 1 + 1/(n+1) + 1/(n+2)² + 1/(n+3)²
< 1 + 1/(n+1) + 1/(n+2)² + 1/(n+3)² + ......
= 1 / (1 - 1/(n+1)) = 1 + 1/n
∴ ( 1 + 1/n )ⁿ⁺¹ > ( 1 + 1/(n+1) )ⁿ⁺²

3)
1 + nx
= (1 + nx) * 1ⁿ⁻¹
By AM ≥ GM ,
≤ ( ((1 + nx) + (n - 1)*1) / n )ⁿ
= (1 + x)ⁿ
∴ (1 + x)ⁿ ≥ 1 + nx

第1條:
http://imageshack.us/f/256/27856651.png/
個idea係prove左P(2^k) is true
然後prove P(n) is true implies P(n-1) is true
咁就conclude 到 P(n) is true for all positive integer n