Chủ đề này có 1 trả lời

[dang123]

$0\leq x,y,z\leq 1$

CM

$(2^{x}+2^{y}+2^{z})(\frac{1}{2^{x}}+\frac{1}{2^{y}}+\frac{1}{2^{z}})\leq \frac{81}{8}$

[phamquanglam]

$0\leq x,y,z\leq 1$

CM

$(2^{x}+2^{y}+2^{z})(\frac{1}{2^{x}}+\frac{1}{2^{y}}+\frac{1}{2^{z}})\leq \frac{81}{8}$

Đặt $a=2^{x};b=2^{y}; c=2^{z}$

Do $(x,y,z)\in \left [ 0;1 \right ]\Rightarrow (a,b,c)\in \left [ 1;2 \right ]$

Nên ta có: $(a-1)(a-2)\leq 0\Leftrightarrow a^{2}+2\leq 3a\Rightarrow a+\frac{2}{a}\leq 3$

Chứng minh tương tự cho $b,c$ ta cũng có: $b+\frac{2}{b}\leq 3; c+\frac{2}{c}\leq 3$

Cộng hết vào:

$\Rightarrow a+b+c+\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\leq 9$

Mặt khác:

Áp dụng AM-GM: $(a+b+c+\frac{2}{}a+\frac{2}{b}+\frac{2}{c})^{2}\geq 4(a+b+c).2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\Rightarrow 8(a+b+c).(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\leq 9^{2}\Rightarrow (a+b+c).(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\leq \frac{81}{8}$

Hay: $(2^{x}+2^{y}+2^{z})+(\frac{1}{2^{x}}+\frac{1}{2^{y}}+\frac{1}{2^{z}})\leq \frac{81}{8}$