求S=24(k^2+1)^2/(3k^2+2)(2k^2+3)的最小值

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求S=24(k^2+1)^2/(3k^2+2)(2k^2+3)的最小值
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求S=24(k^2+1)^2/(3k^2+2)(2k^2+3)的最小值
求S=24(k^2+1)^2/(3k^2+2)(2k^2+3)的最小值

求S=24(k^2+1)^2/(3k^2+2)(2k^2+3)的最小值
如图:


-0.454776,在±1.14103 i 处取得

设k^2=m>=0,则原式=
24(m+1)^2/(3m+2)(2m+3)
=(24m^2+52m+24-4m)/(6m^2+13m+6)
=4-4m/(6m^2+13m+6)
又4m/(6m^2+13m+6)=4/(6m+...

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设k^2=m>=0,则原式=
24(m+1)^2/(3m+2)(2m+3)
=(24m^2+52m+24-4m)/(6m^2+13m+6)
=4-4m/(6m^2+13m+6)
又4m/(6m^2+13m+6)=4/(6m+(6/m)+13)
而6m+6/m>=2√(m*6/m)=12,
所以4m/(6m^2+13m+6)=4/(6m+6/m+13)<=4/25.
所以4-4m/(6m^2+13m+6)>=4-4/25=96/25.
即S的最小值为96/25.

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