Maths Questions

Q1:
(a) t^2/(t-1)=(t+1) + 1/(t-1)= 2+ (t-1)+1/(t-1) >= 2 + 2=4
Without loss of generality, we can suppose the x>=y>=z>=1 for (b)and(c)
(b) x^2/(y-1) + y^2/(z-1) + z^2/(x-1) - [y^2/(y-1) + z^2/(z-1) + x^2/(x-1) ]
=(x^2-y^2)/(y-1)+ (y^2-z^2)/(z-1) - [(x^2-y^2)+(y^2-z^2)]/(x-1)
=(x^2-y^2)[1/(y-1) - 1/(x-1) ] + (y^2-z^2)[ 1/(z-1) - 1/(x-1)]
>= 0 (Since x^2>=y^2>=z^2 and 1/(z-1)>=1/(y-1)>=1/(x-1) )
thus, x^2/(y-1)+..+z^2/(x-1) >= y^2/(y-1)+z^2/(z-1)+x^2/(x-1)
(by (a) ) >= 4+4+4=12
(c) x^4/(y-1)^2+y^4/(z-1)^2+ z^4/(x-1)^2 -[y^4/(y-1)^2+z^4/(z-1)^2+x^4/(x-1)^2]
=(x^4-y^4)/(y-1)^2+ (y^4-z^4)/(z-1)^2 -[(x^4-y^4)+(y^4-z^4)]/(x-1)^2
=(x^4-y^4)[1/(y-1)^2 - 1/(x-1)^2 ] + (y^4-z^4)[ 1/(z-1)^2 - 1/(x-1)^2]
>= 0 (Since x^4>=y^4>=z^4 and 1/(z-1)^2>=1/(y-1)^2>=1/(x-1)^2 )
thus, x^4/(y-1)^2+..+z^4/(x-1)^2 >= y^4/(y-1)^2+z^4/(z-1)^2+x^4/(x-1)^4
(by (a) ) >= 4^2+4^2+4^2=48

Note: the results can be generalized.

Q2:
(1) Obviously 25 | (5^n-25)
(2) 5^n = 1 (mod 4), so 5^n - 25 = 1- 1 (mod 4)
so that 4*25 | (5^n - 25)
(Alternative method)
Let n=2+k, then
5^n-25=25[5^k -1]=25*(5-1)[5^(k-1)+5^(k-2)+...+5+1]
=100*A
so, 5^n - 25 is divisible by 100.

1a) t>1
t-2>-1
(t-1)^2≧0
t^2-4t+4≧0
t^2≧4t-4
t^2 /t-1 ≧ 4 (t-1>0)
b)By(a)x^2 / y-1,y^2 /z-1,z^2 / x-1≧ 4
so,x^2 / y-1 + y^2 /z-1 + 2^2 / x-1 ≧ 12
c)(x^2 / y-1)^2,(y^2 /z-1)^2,(z^2 / x-1)^2≧ 16
so,x^4 / (y-1)^2 + y^4 / (z-1)^2 + z^2 / (x-1)^2 ≧ 48

2)let P(n) be '5^n - 25 is divisible by 100 for n ≧ 2'
for n=2
5^2-25=0 which is divisible by 100
assume P(k) is true,5^k-25=100M
for n=k+1
5^(k+1)-25=5x5^k-25
=5(100M+25)-25=500M+100 which is divisible by 100
P(k+1) is true

2010-02-21 09:33:25 補充：
1a) t>1
t-2>-1
(t-1)^2≧0
t^2-4t+4≧0
t^2≧4t-4
t^2 /t-1 ≧ 4 (t-1>0)
b)By(a)x^2 / y-1,y^2 /z-1,z^2 / x-1≧ 4
so,x^2 / y-1 + y^2 /z-1 + 2^2 / x-1 ≧ 12
c)(x^2 / y-1)^2,(y^2 /z-1)^2,(z^2 / x-1)^2≧ 16
so,x^4 / (y-1)^2 + y^4 / (z-1)^2 + z^2 / (x-1)^2 ≧ 48