Chủ đề này có 5 trả lời

[kitten cute]

1. $x+y\geq 2; x,y> 0$. Tìm min $S=\sqrt[3]{x^{2}+\frac{\sqrt{2}}{y}}+\sqrt[3]{y^{2}+\frac{\sqrt{2}}{x}}$

2. ab+2bc+2ca=13. Tim min $S=21a^{2}+2b^{2}$

3. a+b+c=0. Chung minh $8^{a}+8^{b}+8^{c}\geq 2^{a}+2^{b}+2^{c}$

Bài viết đã được chỉnh sửa nội dung bởi kitten cute: 09-05-2016 - 20:10

[githenhi512]

3. a+b+c=0. Chung minh $8^{a}+8^{b}+8^{c}\geq 2^{a}+2^{b}+2^{c}$

Đ $(2^a;2^b;2^c)\rightarrow (x;y;z)\Rightarrow xyz=2^{a+b+c}=1$

Bđt cần CM $\Leftrightarrow x^3+y^3+z^3\geq x+y+z$

Tc: $x^3+1+1\geq 3x \Rightarrow \sum x^3+6\geq 2\sum x+\sum x\geq 6\sqrt[3]{xyz}+\sum x=6+\sum x\Rightarrow$ đpcm 

Dấu ''='' xr khi x=y=z=1 $\Leftrightarrow$ a=b=c=0

[kitten cute]

Đ $(2^a;2^b;2^c)\rightarrow (x;y;z)\Rightarrow xyz=2^{a+b+c}=1$

Bđt cần CM $\Leftrightarrow x^3+y^3+z^3\geq x+y+z$

Tc: $x^3+1+1\geq 3x \Rightarrow \sum x^3+6\geq 2\sum x+\sum x\geq 6\sqrt[3]{xyz}+\sum x=6+\sum x\Rightarrow$ đpcm 

Dấu ''='' xr khi x=y=z=1 $\Leftrightarrow$ a=b=c=0

minh van chua hieu lam

[githenhi512]

Chỗ nào z bn?

[kitten cute]

Chỗ nào z bn?

Tc: x3+1+1≥3x⇒∑x3+6≥2∑x+∑x≥63√xyz+∑x=6+∑x⇒x3+1+1≥3x⇒∑x3+6≥2∑x+∑x≥6xyz3+∑x=6+∑x⇒ đpcm

[githenhi512]

Tc: x3+1+1≥3x⇒∑x3+6≥2∑x+∑x≥63√xyz+∑x=6+∑x⇒x3+1+1≥3x⇒∑x3+6≥2∑x+∑x≥6xyz3+∑x=6+∑x⇒ đpcm

Áp dụng bđt Cauchy vs x,y,z >0 tc: $x^3+1+1\geqslant 3\sqrt[3]{x^3.1.1}=3x, y^3+1+1\geq \sqrt[3]{y^3.1.1}=3y, z^3+1+1\geq 3z, x+y+z\geq 3\sqrt[3]{xyz}=3\Rightarrow x^3+1+1+y^3+1+1+z^3+1+1\geq 3x+3y+3z\Rightarrow x^3+y^3+z^3+6\geq 2(x+y+z)+(x+y+z)\geq 2.3+(x+y+z)\Rightarrow x^3+y^3+z^3\geq x+y+z$(đpcm)