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[TEX]\blue note:\ \ 2^a.4^a+2^b.4^b+2^c.4^c\ge \frac{\(2^a+2^b+2^c\)\(4^a+4^b+4^c)}{3} [/TEX]

[TEX]3\(2^a.4^a+2^b.4^b+2^c.4^c \)- \(2^a+2^b+2^c\)\(4^a+4^b+4^c)=(2^a-2^b)(4^a-4^b)+(2^b-2^c)(4^b-4^c)+(2^c-2^a)(4^c-4^a)\ge 0[/TEX]

[TEX]\righ 2^a.4^a+2^b.4^b+2^c.4^c\ge \frac{\(2^a+2^b+2^c\)\(4^a+4^b+4^c)}{3} [/TEX]
 
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