gt \Leftrightarrow \frac{1}{x}+ \frac{1}{y}= \frac{1}{x^{2}} +\frac{1}{y^{2}}- \frac{1}{xy}(1) (chia 2 vế cho x^{2}y^{2}) đặt \frac{1}{x}=a; \frac{1}{y}=b khi đó (1) TT a+b=a^{2} +b^{2}-ab=(a+b)^{2}-3ab \geq (a+b)^{2}-\frac{3}{4}(a+b)^{2}=\frac{(a+b)^{2}}{4}$\Rightarrow (a+b)^{2}-4(a+b) \leq0 \Leftrightarrow 0\leq a+b \leq4 A=a^{3} +b^{3}=(a+b)(a^{2}-ab+ b^{2})=(a+b)^{2}\leq 16dấu "="\Leftrightarrow x=y=\frac{1}{2}$
gt
\Leftrightarrow \frac{1}{x}+ \frac{1}{y}= \frac{1}{x^{2}} +\frac{1}{y^{2}}- \frac{1}{xy}(1) (chia 2 vế cho
x^{2}y^{2}) đặt
\frac{1}{x}=a; \frac{1}{y}=b ( a+b >0) khi đó (1) TT
a+b=a^{2} +b^{2}-ab=(a+b)^{2}-3ab \geq (a+b)^{2}-\frac{3}{4}(a+b)^{2}=\frac{(a+b)^{2}}{4}\Rightarrow (a+b)^{2}-4(a+b) \leq0 \Leftrightarrow a+b \leq4(do a+b>0) A=
a^{3} +b^{3}=(a+b)(a^{2}-ab+ b^{2})=(a+b)^{2}\leq 16dấu
"="\Leftrightarrow x=y=\frac{1}{2}