Cho 1\leq a; b; c \leq2
CMR:
[TEX](a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\leq 10[/TEX]
Dấu = xảy ra khi nào?
Ai làm đc thì post lên hộ tớ nhá!
có [TEX]1 \leq a \leq 2 ; 1 \leq c \leq 2 \Rightarrow a \leq 2c ; c \leq 2a \Rightarrow (a-2c)(c-2a) \geq 0.[/TEX]
[TEX]\Rightarrow 2a^2+2b^2 \leq 5ac \Rightarrow \frac{2a^2+2b^2}{ac} \leq 5[/TEX]
[TEX]\Rightarrow \frac{a}{c}+ \frac{c}{a} \leq \frac{5}{2}[/TEX]
Mà [TEX](a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\leq 10 \Leftrightarrow 3+\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{b}+\frac{c}{a}+\frac{c}{a} \leq 10[/TEX]
[TEX]\Leftrightarrow \frac{a}{b}+\frac{b}{a} + \frac{b}{c}+\frac{c}{b} + \frac{c}{b} + \frac{c}{a} + \frac{c}{a} \leq 7.[/TEX]
giả sử [TEX]a \geq b \geq c[/TEX] thì
[TEX]\frac{a}{b} \geq 1 ; \frac{b}{c} \geq 1 \Rightarrow (1-\frac{a}{b})(1-\frac{b}{c}) \geq 0 \Leftrightarrow 1+\frac{a}{c} \geq \frac{a}{b}+\frac{b}{c} (1)[/TEX]
[TEX]\frac{b}{a} \leq 1 ; \frac{c}{b} \leq 1 \Rightarrow (1-\frac{b}{a})(1-\frac{c}{b}) \geq 0 \Rightarrow 1+\frac{c}{a} \geq \frac{b}{a} + \frac{c}{b} (2)[/TEX]
Cộng (1) với (2):
[TEX]\frac{a}{b}+\frac{b}{a}+ \frac{b}{c}+\frac{c}{b} \leq 2+(\frac{a}{c}+\frac{c}{a}) \leq 2+\frac{5}{2}[/TEX]
[TEX]\Rightarrow \frac{a}{b} + \frac{b}{a} + \frac{b}{c} + \frac{c}{b} + \frac{c}{b} + \frac{c}{a} + \frac{c}{a} \leq 2+ \frac{5}{2}+\frac{5}{2}=7[/TEX](đpcm)
Dấu bằng xảy ra khi [TEX]a=b=2, c=1[/TEX];; [TEX]a=2, b=c=1[/TEX] hoặc các hoán vị.