Tìm GTNN

H

huynhbachkhoa23

Dễ mà anh :D

Đặt $P=ab$

$A=\dfrac{(ab+\dfrac{1}{2})(ab+\dfrac{3}{2})}{a^2b^2}=\dfrac{P^2+2P+\dfrac{3}{4}}{P^2}$

Có $0<P \le \dfrac{1}{4}(a+b)^2=\dfrac{1}{16}$

Khảo sát hàm $f(t)=\dfrac{t^2+2t+\dfrac{3}{4}}{t^2}$ trên $(0;\dfrac{1}{16}]$ được $f(t) \ge f(\dfrac{1}{16})=225$

$\text{minA}=225 \leftrightarrow a=b=\dfrac{1}{4}$
 
E

eye_smile

Cách khác:
$A=(\dfrac{1}{a^2}-1)(\dfrac{1}{b^2}-1)=\dfrac{(1-a^2)(1-b^2)}{(ab)^2}=\dfrac{1-(a^2+b^2)+(ab)^2}{(ab)^2}=1+\dfrac{1-(a^2+b^2)}{(ab)^2}=1+\dfrac{\dfrac{3}{4}+(a+b)^2-(a^2+b^2)}{(ab)^2}=1+\dfrac{\dfrac{3}{4}+2ab}{(ab)^2}=1+\dfrac{3}{4(ab)^2}+\dfrac{2}{ab}$

Mà: $\dfrac{1}{2}=a+b \ge 2\sqrt{ab}$

\Leftrightarrow $ab \le \dfrac{1}{16}$

\Rightarrow $A \ge 1+\dfrac{3}{4.\dfrac{1}{16^2}}+\dfrac{2}{\dfrac{1}{16}}=....$

Dấu "=" xảy ra \Leftrightarrow $a=b=\dfrac{1}{4}$
 
You must log in or register to reply here.
Top Bottom