Cho $a,b,c \in [1;3]$ và $a+b+c=6$. Chứng minh bất đẳng thức $60 \le (a+b)(b+c)(c+a) \le 64$

Cần trả +1,000vỏ sò để xem nội dung lời giải này

thanks.... –  Lionel Messi 12-07-16 08:27 PM
dòng thứ 3 từ dưới lên có nhầm 1 tí :)) –  Aerialace 12-07-16 04:35 PM
Áp dụng BĐT Côsi ta được:
$(a+b)(b+c)(c+a)\leq \frac{(2a+2b+2c)^{3}}{27}=64$
Chiều Max:
$\prod(a+b)\leq(\frac{2(a+b+c)}{3})^3=64$
dấu bằng xảy ra khi a=b=c=2
Chiều Min:
Do $a,b,c\in[1;3]$ nên:
.$\prod(a-1)\geq0$$\Leftrightarrow abc-\sum ab +\sum a -1\geq0$
$\Leftrightarrow abc+5\geq \sum ab$
.$\prod(a-3)\leq0$$\Leftrightarrow abc-3\sum ab+9\sum a -27\leq0$
$\Leftrightarrow abc+27 \leq 3\sum  ab $
Do đó, $abc+27\leq3(abc+5)\Rightarrow abc\geq6 ^{(1)}\Rightarrow\sum ab\geq 11^{(2)}$
Mà $\prod(a+b)=(\sum ab)(\sum a)-abc=6\sum ab-abc ^{(3)}$
Since (1),(2) and (3), where are problems in my solution?


mình hiểu, mình hơi lấp tấp. –  duongcscx 12-07-16 05:16 PM
các đánh giá của bạn tuy đúng nhưng không đủ chặt để chứng minh bdt –  Aerialace 12-07-16 04:26 PM

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