就 那條 發(fā)視頻的 一視頻 提到的結(jié)論 之 證明
即
m，n＞0
m2+n2=1
有
a/m^k+b/n^k
≥
(a^(2/(k+2))+b^(2/(k+2)))^((k+2)/2)
之
證明
有
m2+n2=1
即
-akm^(k-1)/m^(2k)
/
m
=
-bkn^(k-1)/n^(2k)
/
n
即
a/m^(k+2)
=
b/n^(k+2)
即
a/b=(m/n)^(k+2)
即
m=(a/b)^(1/(k+2))n
即
m
=
a^(1/(k+2))
/
(a^(2/(k+2))+b^(2/(k+2)))^(1/2)
n
=
b^(1/(k+2))
/
(a^(2/(k+2))+b^(2/(k+2)))^(1/2)
時(shí)
a/m^k+b/n^k
得
最小值
a(a^(2/(k+2))+b^(2/(k+2)))^(k/2)
/
a^(k/(k+2))
+
b(a^(2/(k+2))+b^(2/(k+2)))^(k/2)
/
b^(k/(k+2))
=
a^(2/(k+2))
(a^(2/(k+2))+b^(2/(k+2)))^(k/2)
+
b^(2/(k+2))
(a^(2/(k+2))+b^(2/(k+2)))^(k/2)
=
(a^(2/(k+2))+b^(2/(k+2)))
(a^(2/(k+2))+b^(2/(k+2)))^(k/2)
=
(a^(2/(k+2))+b^(2/(k+2)))^((k+2)/2)
即
m，n＞0
a/m^k+b/n^k
≥
(a^(2/(k+2))+b^(2/(k+2)))^((k+2)/2)
成立
得證
ps.
有關(guān)那條
是那什么
還想立牌坊
骯臟齷齪
腌臜不堪
“秒殺大招”
發(fā)視頻的
無(wú)恥行徑
詳見
與
標(biāo)簽：
