1. Áp dụng BĐT Cauchy ta có:
[tex]\left\{\begin{matrix} (\sqrt{3}-1)^2y^2+x^2\geq 2(\sqrt{3}-1)xy\\ (\sqrt{3}-1)y^2+(\sqrt{3}-1)z^2\geq 2(\sqrt{3}-1)yz\\ (\sqrt{3}-1)^2z^2+x^2\geq 2(\sqrt{3}-1)xz \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{(\sqrt{3}-1)^2}{2}y^2+\frac{1}{2}x^2\geq (\sqrt{3}-1)xy\\ (\sqrt{3}-1)y^2+(\sqrt{3}-1)z^2\geq 2(\sqrt{3}-1)yz\\ \frac{(\sqrt{3}-1)^2}{2}z^2+\frac{1}{2}x^2\geq (\sqrt{3}-1)xz \end{matrix}\right.\Rightarrow x^2+[\frac{(\sqrt{3}-1)^2}{2}+\sqrt{3}-1]y^2+[\frac{(\sqrt{3}-1)^2}{2}+\sqrt{3}-1]z^2\geq (\sqrt{3}-1)(xy+2yz+zx)\Leftrightarrow x^2+y^2+z^2\geq (\sqrt{3}-1)(xy+2yz+zx)\Rightarrow P\geq \sqrt{3}-1[/tex]
Dấu "=" xảy ra khi [tex]x=(\sqrt{3}-1)y=(\sqrt{3}-1)z[/tex]
2. Ta có:[tex]A=x^2+y^2-xy=(x+y)^2-3xy=16x^2y^2-3xy[/tex]
Vì [tex]4xy=x+y\geq 2\sqrt{xy}\Rightarrow 2\sqrt{xy}(2\sqrt{xy}-1)\geq 0\Rightarrow 2\sqrt{xy}\geq 1\Rightarrow xy\geq \frac{1}{4}[/tex]
Lại có: [tex](1-x)(1-y)\geq 0\Rightarrow xy+1\geq x+y=4xy\Rightarrow xy\leq \frac{1}{3}[/tex]
[tex]A=x^2+y^2-xy=(x-y)^2+xy\geq 0+\frac{1}{4}=\frac{1}{4}[/tex]
Dấu "=" xảy ra khi [tex]x=y=\frac{1}{2}[/tex]
Mà [tex]A=16x^2y^2-3xy=7x^2y^2+3xy(3xy-1)\leq 7.(xy)^2+0=7.\frac{1}{9}=\frac{7}{9}[/tex]
Dấu "=" xảy ra khi [tex]x=\frac{1}{3},y=1 hoặc x=1,y=\frac{1}{3}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
