$\sum_{cyc} (x+y).\sqrt{(z+x)(z+y)} \ge 4(xy+yz+xz) \forall x,y,z>0$
$\sum_{cyc} (x+y).\sqrt{(z+x)(z+y)} \ge 4(xy+yz+xz)$
Started By I Love MC, 23-08-2014 - 19:24
#1
Posted 23-08-2014 - 19:24
#2
Posted 23-08-2014 - 20:24
$\sum_{cyc} (x+y).\sqrt{(z+x)(z+y)} \ge 4(xy+yz+xz) \forall x,y,z>0$
Theo AM-GM kết hợp với bdt $(a+b)(b+c)(c+a)\geq \frac{8}{9}(a+b+c)(ab+bc+ac)$
$a+b+c\geq \sqrt{3(ab+bc+ac)}$
$= > \sum (x+y)\sqrt{(y+z)(x+z)}=\sqrt{\prod (x+y)}(\sum \sqrt{x+y})\geq \sqrt{\frac{8}{9}(\sum x)(\sum xy)}.3\sqrt[6]{\prod (x+y)}\geq \sqrt{\frac{8}{9}\sqrt{3\sum xy}.(\sum xy)}.3\sqrt[6]{\frac{8}{9}(\sum x)(\sum xy)}\geq \sqrt{\frac{8}{9}.3\sqrt{\sum xy}.(\sum xy)}.3\sqrt[6]{\frac{8}{9}(\sum xy).\sqrt{3\sum xy}}=4\sum xy$
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