Ответ:

[tex](x+\frac{1}{y})(y+\frac{1}{x})\geq 4[/tex]

Объяснение:

формула:

[tex](a-b)^2\ge0\\\\a^2-2ab+b^2\ge0\ \ \ |+4ab\\\\a^2+2ab+b^2\ge 4xy\\\\(a+b)^2\ge 4ab\ \ \ |\sqrt{}\\\\a+b\ge 2\sqrt{ab}[/tex]

[tex](x+\frac{1}{y})(y+\frac{1}{x})\geq 2\sqrt{x\cdot\frac{1}{y}}\cdot 2\sqrt{y\cdot\frac{1}{x}}\\\\(x+\frac{1}{y})(y+\frac{1}{x})\geq 4\sqrt{\frac{x}{y}}\cdot\sqrt{\frac{y}{x}}\\\\(x+\frac{1}{y})(y+\frac{1}{x})\geq 4\sqrt{\frac{x}{y}\cdot\frac{y}{x}}\\\\(x+\frac{1}{y})(y+\frac{1}{x})\geq 4\sqrt{1}\\\\(x+\frac{1}{y})(y+\frac{1}{x})\geq 4[/tex]