Question

\(\text{Cho x,y}\in R\text{ thỏa mãn }x^2+y^2=4.\text{Tìm Max}\)

\(A=\frac{xy}{x+y+2}\)

Answer 1

Lời giải:

Đặt $xy=a; x+y=b$ thì ta có: \(\left\{\begin{matrix} b^2-2a=4\\ b^2\geq 4a\end{matrix}\right.\)

$A=\frac{xy}{x+y+2}=\frac{a}{b+2}=\frac{b^2-4}{2(b+2)}=\frac{b-2}{2}$
Từ $b^2\geq 4a$. Thay $4a=2(b^2-4)$ có:

$b^2\geq 2(b^2-4)$

$\Leftrightarrow b^2\leq 8\Rightarrow b\leq 2\sqrt{2}$

Do đó: $A=\frac{b-2}{2}\leq \frac{2\sqrt{2}-2}{2}=\sqrt{2}-1$

Vậy $A_{\max}=\sqrt{2}-1$