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L

leminhnghia1

Giải:

Theo bđt Cô-si ta có:

[TEX]1 \ \geq \ a+\frac{1}{b} \ \geq \ 2\sqrt{\frac{a}{b}}[/TEX]

[TEX]\Rightarrow \ \frac{a}{b} \ \leq \ \frac{1}{4}[/TEX]

Ta có: [TEX]\frac{a}{b}+\frac{b}{a} \ = \ ( \ \frac{16a}{b}+\frac{b}{a} \ )-\frac{15a}{b} \ \geq \ 2\sqrt{16} \ - \ \frac{15a}{b} \ \geq \ 8 \ - \ \frac{15}{4}=\frac{17}{4}[/TEX]

GTNN Min $C \ = \ \frac{17}{4}$ có khi: $a=\frac{1}{2}$ , $b=2$
 
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