Câu hỏi của Easylove

Question

Cho x, y, z dương thoả mãn

$\sqrt{x}+\sqrt{y}+\sqrt{z}=3$
tìm GTNN của $M=\sqrt{x+\frac{1}{x}}+\sqrt{y+\frac{1}{y}}+\sqrt{z+\frac{1}{z}}$

Nguyễn Việt Lâm · Accepted Answer

Đặt $\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)=\left(a;b;c\right)\Rightarrow a+b+c=3$

$M=\sqrt{a^2+\frac{1}{a^2}}+\sqrt{b^2+\frac{1}{b^2}}+\sqrt{c^2+\frac{1}{c^2}}$

$M\ge\sqrt{\left(a+b+c\right)^2+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}$

$M\ge\sqrt{\left(a+b+c\right)^2+\frac{81}{\left(a+b+c\right)^2}}$

$M\ge\sqrt{2\sqrt{\frac{81\left(a+b+c\right)^2}{\left(a+b+c\right)^2}}}=3\sqrt{2}$

$M_{min}=3\sqrt{2}$ khi $a=b=c=1$Câu trả lời: Đặt $\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)=\left(a;b;c\right)\Rightarrow a+b+c=3$

$M=\sqrt{a^2+\frac{1}{a^2}}+\sqrt{b^2+\frac{1}{b^2}}+\sqrt{c^2+\frac{1}{c^2}}$

$M\ge\sqrt{\left(a+b+c\right)^2+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}$

$M\ge\sqrt{\left(a+b+c\right)^2+\frac{81}{\left(a+b+c\right)^2}}$

$M\ge\sqrt{2\sqrt{\frac{81\left(a+b+c\right)^2}{\left(a+b+c\right)^2}}}=3\sqrt{2}$

$M_{min}=3\sqrt{2}$ khi $a=b=c=1$

Chương 4: BẤT ĐẲNG THỨC, BẤT PHƯƠNG TRÌNH