Chủ đề này có 4 trả lời

[Khoa Linh]

Cho a, b, c >0 thỏa mãn a+b+c=3. Chứng minh BĐT sau:

 

 

$(a+b)(b+c)(c+a)\geq (a+bc)(b+ca)(c+ab)$

[DOTOANNANG]

$$(a+b)(b+c)(c+a)\geq (a+bc)(b+ca)(c+ab)$$

$$8abc\leq \left ( a+ b \right )\left ( b+ c \right )\left ( c+ a \right )$$

$$\Leftrightarrow 8\left ( a+ b \right )\left ( b+ c \right )\left ( c+ a \right )+ 8abc\leq 9\left ( a+ b \right )\left ( b+ c \right )\left ( c+ a \right )$$

$$\Leftrightarrow 9\left ( a+ b \right )\left ( b+ c \right )\left ( c+ a \right )\geq 8\left ( a+ b+ c \right )\left ( ab+ bc+ ca \right )$$

$$= 24\left ( ab+ bc+ ca \right )\geq \frac{\left ( 3+ bc+ ca+ ab \right )^{3}}{3}= \frac{\left ( a+ b+ c+ bc+ ca+ ab \right )^{3}}{3}\geq 9\left ( a+ bc \right )\left ( b+ ca \right )\left ( c+ ab \right )$$

[moriran]

Đoạn $24\left ( ab+ bc+ ca \right )\geq \frac{\left ( 3+ bc+ ca+ ab \right )^{3}}{3}$ cm sao vậy bạn?

[DOTOANNANG]

$$0\leq ab+ bc+ ca\leq \frac{\left ( a+ b+ c \right )^{2}}{3}= 3$$

$$24\left ( ab+ bc+ ca \right )\geq \frac{\left ( 3+ bc+ ca+ ab \right )^{3}}{3}$$

$$\Leftrightarrow \frac{1}{3}\left ( 3- \left [ab+ bc+ ca \right ] \right )\left ( \left [ ab+ bc+ ca \right ]^{2}+ 12\left [ ab+ bc+ ca \right ]- 9 \right )\geq 0$$

 

[Khoa Linh]

Mình đã nghĩ ra cách giải khác 

$(a+b)(c+1)=(ac+b)+(bc+a)\geq 2\sqrt{(ac+b)(bc+a)}$

Tương tự 

$\Rightarrow (a+1)(b+1)(c+1)(a+b)(b+c)(c+a)\geq 8(ab+c)(bc+a)(ca+b)$

Mà theo AM-GM ta có:

$(a+1)(b+1)(c+1)\leq \left ( \frac{a+b+c+3}{3} \right )^3=8$

Suy ra ĐPCM 

Bài viết đã được chỉnh sửa nội dung bởi Khoa Linh: 28-02-2018 - 20:34