[Toán 8]BDT

C

congchuaanhsang

Áp dụng BĐT Cauchy ta có:

16=$[a+(b+c)]^2$\geq$4a(b+c)$\Leftrightarrow4\geq$a(b+c)$

\Leftrightarrow$4(b+c)$\geq$a(b+c)^2$ (vì b+c>0)

\Rightarrow$4(b+c)$\geq$4abc$ (vì $(b+c)^2$\geq$4bc$)\Leftrightarrowb+c\geqabc

Dấu "=" xảy ra \Leftrightarrow a=2 ; b=c=1
 
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