Cho x,y,z là các số dương. Chứng minh rằng [tex](\frac{x}{y}+\frac{y}{z}+\frac{z}{x})^2-(x+y+z).(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 0[/tex]
[tex]BDT\Leftrightarrow \frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}+\frac{x}{z}+\frac{z}{y}+\frac{y}{x}\geq 3+\frac{x}{y}+\frac{y}{z}+\frac{z}{x}[/tex]
Đặt [tex](\frac{x}{z};\frac{z}{y};\frac{y}{x})=(a;b;c)(>0)\Rightarrow abc=1[/tex]
Ta cần chứng minh [tex]\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+a+b+c\geq 3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}[/tex]
Có [tex]\frac{1}{a^2}+1+\frac{1}{b^2}+1+\frac{1}{c^2}+1\geq^{AM-GM} 2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})[/tex]
[tex]\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+3+a+b+c\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+a+b+c)\geq^{AM-GM}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+6[/tex]
[tex]\Rightarrow VT\geq VP(đpcm)[/tex]
Dấu = xảy ra khi [tex]x=y=z[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
