Question

Довести нерівність (xy+16)(x+y)≥16xy якщо x ≥ 0, y ≥ 0.​

Answer 1

Ответ:

$(xy+16)(x+y)\geq16xy$

Объяснение:

формула:

$(x-y)^2\ge0\\\\x^2-2xy+y^2\ge0\ \ \ |+4xy\\\\x^2+2xy+y^2\ge 4xy\\\\(x+y)^2\ge 4xy\ \ \ |\sqrt{}\\\\x+y\ge 2\sqrt{xy}$

$(xy+16)(x+y)\geq2\sqrt{xy\cdot 16}\cdot2\sqrt{x\cdot y}\\\\(xy+16)(x+y)\geq4\sqrt{16xy}\cdot\sqrt{xy}\\\\(xy+16)(x+y)\geq4\sqrt{16xy\cdot xy}\\\\(xy+16)(x+y)\geq4\sqrt{16x^2y^2}\\\\(xy+16)(x+y)\geq4\sqrt{(4xy)^2}\\\\(xy+16)(x+y)\geq4\cdot|4xy|\\\\(xy+16)(x+y)\geq4\cdot 4xy\\\\(xy+16)(x+y)\geq16xy$