Chủ đề này có 4 trả lời

[hien2000a]

a)Cho x,y,z,a,b,c>0 .CMR:

$\sqrt[3]{abc}+\sqrt[3]{xyz}\leq \sqrt[3]{(a+x)(b+y)(c+z)}$

b) từ a) suy ra:

$\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}\leq 2\sqrt[3]{3}$

[githenhi512]

a)Cho x,y,z,a,b,c>0 .CMR:

$\sqrt[3]{abc}+\sqrt[3]{xyz}\leq \sqrt[3]{(a+x)(b+y)(c+z)}$

b) từ a) suy ra:

$\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}\leq 2\sqrt[3]{3}$

a. $\frac{\sqrt[3]{abc}+\sqrt[3]{xyz}}{\sqrt[3]{(a+x)(b+y)(c+z)}}$

$=\sqrt[3]{\frac{a}{a+x}.\frac{b}{b+y}.\frac{c}{c+z}}+\sqrt[3]{\frac{x}{a+x}.\frac{y}{b+y}.\frac{z}{c+z}}.$

$\leq \frac{1}{3}(\frac{a}{a+x}+\frac{b}{b+y}+\frac{c}{c+z}+\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{z+c})=1$

(đpcm)

b. $VT=\sqrt[3]{\sqrt[3]{3}.(\sqrt[3]{3^{2}}+1).1}+\sqrt[3]{\sqrt[3]{3}.(\sqrt[3]{3^{2}}-1).1}.$

            $\leq \sqrt[3]{2.\sqrt[3]{3}.2\sqrt[3]{3^{2}}.2}=VP.$(đpcm)

Bài viết đã được chỉnh sửa nội dung bởi githenhi512: 21-04-2016 - 22:54

[hien2000a]

a. $\frac{\sqrt[3]{abc}+\sqrt[3]{xyz}}{\sqrt[3]{(a+x)(b+y)(c+z)}}$

$=\sqrt[3]{\frac{a}{a+x}.\frac{b}{b+y}.\frac{c}{c+z}}+\sqrt[3]{\frac{x}{a+x}.\frac{y}{b+y}.\frac{z}{c+z}}.$

$\leq \frac{1}{3}(\frac{a}{a+x}+\frac{b}{b+y}+\frac{c}{c+z}+\frac{x}{a+x}+\frac{y}{b+y}+\frac{z}{z+c})=1$

(đpcm)

b. $VT=\sqrt[3]{\sqrt[3]{3}.(\sqrt[3]{3^{2}}+1).1}+\sqrt[3]{\sqrt[3]{3}.(\sqrt[3]{3^{2}}-1).1}.$

            $\leq \sqrt[3]{2.\sqrt[3]{3}.2\sqrt[3]{3^{2}}.2}=VP

bạn có thể chứng minh BĐT bạn dung ko?

[chaubee2001]

a)Cho x,y,z,a,b,c>0 .CMR:
$\sqrt[3]{abc}+\sqrt[3]{xyz}\leq \sqrt[3]{(a+x)(b+y)(c+z)}$
b) từ a) suy ra:
$\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}\leq 2\sqrt[3]{3}$

cách khác cho bạn nè
$BĐT \Leftrightarrow abc+xyz+3\sqrt[3]{abcxyz}(\sqrt[3]{abc}+\sqrt[3]{xyz})\leq (a+x)(b+y)(c+z)$
$\Leftrightarrow abc+xyz+3\sqrt[3]{abcxyz}(\sqrt[3]{abc}+\sqrt[3]{xyz}) \leq abc+xyz+abz+acy+ayz+bcx+bxz+cxy$
$\Leftrightarrow 3\sqrt[3]{abcxyz}(\sqrt[3]{abc}+\sqrt[3]{xyz}) \leq abz+acy+ayz+bcx+bxz+cxy$
Mà: $AM-GM: abz+acy+bcz \geq 3\sqrt[3]{a^2b^2c^2xyz}$
$bxz+cxy+ayz \geq 3\sqrt[3]{x^2y^2z^2abc}$
Suy ra $abz+acy+bcz+bxz+cxy+ayz \geq 3\sqrt[3]{abcxyz}(\sqrt[3]{abc}+\sqrt[3]{xyz})$
Ý b thỳ khỏi phải nói

[githenhi512]

bạn có thể chứng minh BĐT bạn dung ko?

Bđt Cauchy: $\sqrt[3]{abc}\leq a+b+c$(a,b,c>0)

Dấu ''='' xr khi a=b=c>0