Chủ đề này có 3 trả lời

[HoangPhat94]

Cho 3 số a,b,c dương. Chứng minh rằng

 

$(a^{3}+b^{3}+c^{3})(\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}})\geq \frac{3}{2}(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b})$

Bài viết đã được chỉnh sửa nội dung bởi HoangPhat94: 26-11-2013 - 23:45

[kfcchicken98]

có $\frac{a^{3}}{b^{3}}+1+1\geq \frac{3a}{b}$

$\frac{a^{3}}{c^{3}}+1+1\geq \frac{3a}{c}$

$\frac{b^{3}}{a^{3}}+1+1\geq \frac{3b}{a}$

$\frac{b^{3}}{c^{3}}+1+1\geq \frac{3b}{c}$

$\frac{c^{3}}{a^{3}}+1+1\geq \frac{3c}{a}$

$\frac{c^{3}}{b^{3}}+1+1\geq \frac{3c}{b}$

suy ra $(a^{3}+b^{3}+c^{3})(\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}})+9\geq 3(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b})$

do $\frac{3}{2}(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}) =\frac{3}{2}(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a}{b})\geq \frac{3}{2}(3+3)=9$

suy ra $(a^{3}+b^{3}+c^{3})(\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}})\geq \frac{3}{2}(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b})$

[HoangPhat94]

có $\frac{a^{3}}{b^{3}}+1+1\geq \frac{3a}{b}$

$\frac{a^{3}}{c^{3}}+1+1\geq \frac{3a}{c}$

$\frac{b^{3}}{a^{3}}+1+1\geq \frac{3b}{a}$

$\frac{b^{3}}{c^{3}}+1+1\geq \frac{3b}{c}$

$\frac{c^{3}}{a^{3}}+1+1\geq \frac{3c}{a}$

$\frac{c^{3}}{b^{3}}+1+1\geq \frac{3c}{b}$

suy ra $(a^{3}+b^{3}+c^{3})(\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}})+9\geq 3(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b})$

do $\frac{3}{2}(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}) =\frac{3}{2}(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a}{b})\geq \frac{3}{2}(3+3)=9$

suy ra $(a^{3}+b^{3}+c^{3})(\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}})\geq \frac{3}{2}(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b})$

Hình như bạn sai, cả 2 vế đều lớn hơn hoặc bằng 9 nk

[TonnyMon97]

Hình như bạn sai, cả 2 vế đều lớn hơn hoặc bằng 9 nk

Đó là cộng vế theo vế  2 BĐT bạn ơi... đúng rồi