~~~代數最值習題~~~

Let f(x)=x^6-x^5-2x^4+x^3+x^2+1

f(x)
=(x^2-1)(x^4-x^3-x^2)+1
=(x^2-1)(x^2-x-1)(x^2)+1
=(x^2-1)(x^2-x+(1/2)^2-(1/2)^2-1)(x^2)+1
=(x^2-1)((x-1/2)^2-3/4)(x^2)+1

if x<0
x^2-1>=0
(x-1/2)^2-3/4>-1/2
x^2>=0

if x^2-1 and x^2 are also equal to 0
the smallest value of (x^2-1)((x-1/2)^2-3/4)(x^2) is -1/2
So the smallest value of f(x) is 1>0

if x=0
x^2-1=-1
x^2-x-1=-1
x^2=0
the value is (-1)(-1)(0)+1=1>0

if x>0
x^2-1>=0
((x-1/2)^2-3/4)>-1/2

So the smallest value of f(x) is 1>0

So x^6-x^5-2x^4+x^3+x^2+1>0

2010-10-24 10:45:13 補充：
actually simplify x^6-x^5-2^4+x^3+x^2+1
and find the properties of each brackets
but可能有誤....有咩疑問再問la~~~~

根據之前的分析可知x≤1時都對(不過我刪了-.-)，補一下x>1時的證明

令x=1+t，t>0，原式可等價成t^6+5t^5+8t^4+3t^3+1>3t^2+2t

由算幾不等式可得3t^3 + (11/32) + (11/32) ≥ (27*121*3/1024)^(1/3) > 2t

且 8t^4 + (5/16) > 3t^2 [判別式<0]

兩式相加得 8t^4+3t^3+1>3t^2+2t，因此t^6+5t^5+8t^4+3t^3+1>3t^2+2t 成立。

The polynomial can be factoried as
(x^2 +2.01827878245301x+1.12383083311722)
(x^2 -0.174212422483314x+0.437208164061247)
(x^2 -2.8440663599697+2.0352173387035)

真係有D疑問 ,

(x^2-1)((x-1/2)^2-3/4)(x^2)+1 好似計錯 ,

if x<0
x^2-1>=0

呢句好似不對,
-0.5<0
(-0.5)^2 - 1 < 0

暫時看

2010-11-04 00:02:14 補充：
是啊!你還真細心^^