........................BĐT............................

Question

cho 3 số a,b,c dương.CMR:

$\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}\leq \sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}$

tran85295 · Answer 1 · 3/15/2016 1:57:14 PM

Bình chọn tăng 12 Bình chọn giảm

Đặt

$\sqrt[6]{\frac ab}=x,\sqrt[6]{\frac bc}=y,\sqrt[6]{\frac ca}=z(x,y,z >0)$

Áp dụng bđt bunhia

$VP=\sqrt[3]{3(a+b+c)(\frac 1b + \frac 1c +\frac 1a)} \ge \sqrt[3]{3(\sqrt{ \frac ab}+\sqrt{\frac bc}+\sqrt{\frac ca})^2}$

$=\sqrt[3]{3(x^3+y^3+z^3)^2}$

Lại có

$VT=x^2+y^2+z^2$

Nên ta chứng minh

$x^2+y^2+z^2 \le \sqrt[3]{3(x^3+y^3+z^3)^2}$

$\Leftrightarrow (1+1+1)(x^3+y^3+z^3)(x^3+y^3+z^3) \ge (x^2+y^2+z^2)^3$ (đúng theo bđt holder)

Vậy ta có đpcm,

$"="\Leftrightarrow a=b=c$

Trả lời 15-03-16 01:57 PM

tran85295
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bạn ko hiểu chỗ nào ? – tran85295 15-03-16 05:40 PM

mình ko hiểu. còn cách nào khác ko – Lionel Messi 15-03-16 02:20 PM

Nguyễn Nhung · Answer 2 · 6/12/2016 9:18:04 AM

Bình chọn tăng 7 Bình chọn giảm

C2)

bđt

$\Leftrightarrow \frac{\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}} \leq1$

$\Leftrightarrow \sum \frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+ \frac{1}{b} +\frac{1}{c})}}\leq1$

ta có

$\frac{\sqrt[3]{\frac{a}{b}}}{\sqrt[3]{3(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}} =\sqrt[3]{\frac{1}{3}.\frac{a}{a+b+c}.\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}}$