Cho 3 số thực a,b,c dương. CMR:
$\frac{6}{\sqrt{x}}+\frac{11}{\sqrt{b}}+\frac{7}{\sqrt{c}}\geqslant 8(\frac{3}{\sqrt{a+3b}}+\frac{2}{\sqrt{b+3c}}+\frac{1}{\sqrt{c+3a}})$
Bài viết đã được chỉnh sửa nội dung bởi hangnguyen2003: 16-12-2018 - 06:34
$\frac{6}{\sqrt{x}}+\frac{11}{\sqrt{b}}+\frac{7}{\sqrt{c}}$
Bắt đầu bởi hangnguyen2003, 16-12-2018 - 06:31
#1
Đã gửi 16-12-2018 - 06:31
hangnguyen2003
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Binh nhất
It doesn't matter if you're the slowest kid in gym class or the fastest man alive. Every one of us is running, being alive is running, running from something, running to something or someone. And no matter how fast you are. There's some things you can't outrun. Some things always manage to catch up to you.
___ THE FLASH ___
#2
Đã gửi 16-12-2018 - 08:44
Arthur Pendragon
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Trung sĩ
$VT=(\frac{3}{\sqrt{a}}+\frac{27}{3\sqrt{b}})+(\frac{2}{\sqrt{b}}+\frac{18}{3\sqrt{c}})+(\frac{1}{\sqrt{c}}+\frac{9}{3\sqrt{a}})$
$\geq \frac{48}{\sqrt{a}+3\sqrt{b}}+\frac{32}{\sqrt{b}+3\sqrt{c}}+\frac{16}{\sqrt{c}+3\sqrt{a}}$
$\geq \frac{48}{2\sqrt{a+3b}}+ \frac{32}{2\sqrt{b+3c}}+ \frac{16}{2\sqrt{c+3a}}=VP$
"WHEN YOU HAVE ELIMINATED THE IMPOSSIBLE, WHATEVER REMAINS, HOWEVER IMPROBABLE, MUST BE THE TRUTH"
-SHERLOCK HOLMES-
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