Đáp án:
Giải thích các bước giải:
ta có: a+b+c=1
⇒$(a+b+c)^{2}$≥4a(b+c)
ta lại có: $(b+c)^{2}$≥4bc
nhân từng vế:
⇒$(a+b+c)^{2}$.$(b+c)^{2}$≥4a(b+c).4bc
⇒b+c≥16abc (ĐPCM)
`a+b+c=1` `(a,b,c>0)`
`<=>a+bc=1-b-c+bc`
`<=>(1-b)(1-c)=(c+a)(a+b)`
`b+c>=16abc`
Áp dụng `BĐT` Cô-si , ta có : `(a+b+c)^2>=4a(b+c);(b+c)^2>=4bc`
Nhân từng vế : `(a+b+c)^2 (b+c)^2>=4a(b+c).4bc=>b+c>=16ab` `->(đpcm)`
Dấu ''='' xảy ra khi : $\begin{cases}a=\dfrac{1}{2}\\b=c=\dfrac{1}{4}\end{cases}$
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