Question

даю 50 балів
завдання на фото

Answer 1

Ответ:

$(a+1)(b+5)(c+2)(d+1)\ge 160\sqrt{abcd}$

Объяснение:

формула:

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

$(a+1)(b+5)(c+2)(d+1)\ge2\sqrt{a\cdot 1}\cdot 2\sqrt{b\cdot 5}\cdot 2\sqrt{c\cdot 2}\cdot 2\sqrt{d\cdot 10}\\\\(a+1)(b+5)(c+2)(d+1)\ge 16\sqrt{a}\cdot \sqrt{5b}\cdot \sqrt{2c}\cdot \sqrt{10d}\\\\(a+1)(b+5)(c+2)(d+1)\ge 16\sqrt{a\cdot 5b \cdot2c\cdot 10d}\\\\(a+1)(b+5)(c+2)(d+1)\ge 16\sqrt{100abcd}\\\\(a+1)(b+5)(c+2)(d+1)\ge 16\cdot 10\sqrt{abcd}\\\\(a+1)(b+5)(c+2)(d+1)\ge 160\sqrt{abcd}$