Tìm giá trị nhỏ nhất của hàm số f(x,y)=asin^4(x)+bcos^4y/csin^2(x)+dcos^2y

* Hướng dẫn giải

Ta có 

$c+d=c\left({\sin }^2\left(x\right)+{\cos }^2\left(x\right)\right)+d\left({\sin }^2\left(y\right)+{\cos }^2\left(y\right)\right)$

Do đó ( O;R )

$$\begin{gathered} \left(c+d\right)f_1=\left[\left(c{\sin }^2\left(x\right)+d{\cos }^2\left(y\right)\right)+\left(c{\cos }^2\left(x\right)+d{\sin }^2\left(x\right)\right)\right] \\ \left[\frac{{\sin }^4\left(x\right)}{c{\sin }^2\left(x\right)+d{\cos }^2\left(y\right)}+\frac{{\cos }^4\left(x\right)}{c{\cos }^2\left(x\right)+d{\sin }^2\left(y\right)}\right] \\ \ge \left(\sqrt{c{\sin }^2\left(x\right)+d{\cos }^2\left(y\right)}-\frac{{\sin }^2\left(x\right)}{\sqrt{c{\sin }^2\left(x\right)+d{\cos }^2\left(y\right)}}\right) \\ +\sqrt{c{\cos }^2\left(x\right)+d{\sin }^2\left(y\right)}-\frac{{\cos }^2\left(x\right)}{\sqrt{c{\cos }^2\left(x\right)+d{\sin }^2\left(y\right)}} \\ =1\Rightarrow f_1\ge \frac{1}{c+d} \end{gathered}$$

Tương tự $f_2\ge \frac{1}{c+d}$. Vậy $f\left(x,y\right)=a{f}_1+b{f}_2\ge \frac{a+b}{c+d}$

Đáp án A