Cho $a,b,c\ge1; abc\ge8$, CMR: $$\sqrt{a^2-1}+\sqrt{b^2-1}+\sqrt{c^2-1}\ge 3\sqrt3$$
Cho $abc\ge8$, CMR $\sqrt{a^2-1}+\sqrt{b^2-1}+\sqrt{c^2-1}\ge 3\sqrt3$
Started By LzuTao, 02-09-2015 - 16:11
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Posted 05-09-2015 - 15:44
longatk08
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Cho $a,b,c\ge1; abc\ge8$, CMR: $$\sqrt{a^2-1}+\sqrt{b^2-1}+\sqrt{c^2-1}\ge 3\sqrt3$$
Thực tế ta chỉ cần chứng minh BĐT khi $abc=8$ khi đó chứng minh BĐT sau:
$\sqrt{a^2-1}+\sqrt{b^2-1}\geq 2\sqrt{ab-1}$ giả sử $c=min${$a,b,c$}
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