Ответ:
[tex](2+a)(2+b)\geq 8\sqrt{ab}[/tex]
Объяснение:
формула:
[tex](x-y)^2\ge0\\\\x^2-2xy+y^2\ge0\ \ \ |+4xy\\\\x^2+2xy+y^2\ge 4xy\\\\(x+y)^2\ge 4xy\ \ \ |\sqrt{}\\\\x+y\ge 2\sqrt{xy}[/tex]
[tex](2+a)(2+b)\geq 2\sqrt{2\cdot a}\cdot 2\sqrt{2\cdot b}\\\\(2+a)(2+b)\geq 4\sqrt{2a}\cdot \sqrt{2b}\\\\(2+a)(2+b)\geq 4\sqrt{2a\cdot 2b}\\\\(2+a)(2+b)\geq 4\sqrt{4ab}\\\\(2+a)(2+b)\geq 4\cdot 2\sqrt{ab}\\\\(2+a)(2+b)\geq 8\sqrt{ab}[/tex]
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Answers & Comments
Ответ:
[tex](2+a)(2+b)\geq 8\sqrt{ab}[/tex]
Объяснение:
формула:
[tex](x-y)^2\ge0\\\\x^2-2xy+y^2\ge0\ \ \ |+4xy\\\\x^2+2xy+y^2\ge 4xy\\\\(x+y)^2\ge 4xy\ \ \ |\sqrt{}\\\\x+y\ge 2\sqrt{xy}[/tex]
[tex](2+a)(2+b)\geq 2\sqrt{2\cdot a}\cdot 2\sqrt{2\cdot b}\\\\(2+a)(2+b)\geq 4\sqrt{2a}\cdot \sqrt{2b}\\\\(2+a)(2+b)\geq 4\sqrt{2a\cdot 2b}\\\\(2+a)(2+b)\geq 4\sqrt{4ab}\\\\(2+a)(2+b)\geq 4\cdot 2\sqrt{ab}\\\\(2+a)(2+b)\geq 8\sqrt{ab}[/tex]