Assume that a>=b>=c
We will prove ab+bc+ca<=(a^8+b^8+c^8)/a^2b^2c^2 by following steps:
ab+bc+ca<=a^2+b^2+c^2 (Rearrangement Inequality)
a^4b^2c^2+a^2b^4c^2+a^2b^2c^4<=a^6b^2+b^6c^2+c^6a^2
<=a^8+b^8+c^8 (Random sum>=Reverse sum)
Furthermore, the equality holds when a=b=c.
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由 宇智波鼬 於 星期一 四月 23, 2007 9:54 pm
[數學]不等式.
由 skywalker 於 星期五 四月 20, 2007 10:34 pm
設a,b,c皆為正實數試證: