a) Theo BĐT AM-GM ta có:
[tex]\frac{x}{yz}+\frac{y}{zx}\geq 2\sqrt{\frac{x}{yz}.\frac{y}{zx}}=\frac{2}{z}[/tex]
Tương tự...
Suy ra [tex]\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}\geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}[/tex]
Suy ra [tex]\frac{x^2}{2}+\frac{y^2}{2}+\frac{z^2}{2}+\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}\\\geq \frac{x^2}{2}+\frac{y^2}{2}+\frac{z^2}{2}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\\=\left ( \frac{x^2}{2}+\frac{1}{2x}+\frac{1}{2x} \right )+\left ( \frac{y^2}{2}+\frac{1}{2y}+\frac{1}{2y} \right )+\left ( \frac{z^2}{2}+\frac{1}{2z}+\frac{1}{2z}\right )\\\geq 3\sqrt[3]{\frac{x^2}{2}.\frac{1}{2x}.\frac{1}{2x}}+3\sqrt[3]{\frac{y^2}{2}.\frac{1}{2y}.\frac{1}{2y}}+3\sqrt[3]{\frac{z^2}{2}.\frac{1}{2z}.\frac{1}{2z}}\\=\frac{9}{2}[/tex]
Dấu = xảy ra khi [tex]x=y=z=1[/tex]
b) [tex]P=\frac{1}{x^2+y^2}+\frac{1}{xy}+xy\\= \frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+xy\\\geq \frac{4}{x^2+y^2+2xy}+\frac{1}{2xy}+8xy-7xy\\\geq \frac{4}{(x+y)^2}+2\sqrt{\frac{1}{2xy}.8xy}-7.\frac{(x+y)^2}{4}\\\geq \frac{4}{1^2}+4-7.\frac{1^2}{4}\\=\frac{25}{4}[/tex]
Dấu = xảy ra khi [tex]x=y=\frac{1}{2}[/tex]
c) [tex]4ab=a+b\geq 2\sqrt{ab}\Leftrightarrow 2ab-\sqrt{ab}\geq 0\Leftrightarrow \sqrt{ab}(2\sqrt{ab}-1)\geq 0\Leftrightarrow ab\geq \frac{1}{4}(vi:ab>0)[/tex]
[tex]\frac{a}{4b^2+1}+\frac{b}{4a^2+1}\\=\frac{a^2}{4ab^2+a}+\frac{b^2}{4a^2b+b}\\\geq \frac{(a+b)^2}{4ab^2+a+4a^2b+b}\\=\frac{(a+b)^2}{(a+b)(4ab+1)}\\=\frac{a+b}{4ab+1}\\=\frac{4ab}{4ab+1}\\=1-\frac{1}{4ab+1}\\\geq 1-\frac{1}{4.\frac{1}{4}+1}\\=\frac{1}{2}[/tex]
Dấu = xảy ra khi [tex]a=b=\frac{1}{2}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
