$\sqrt{x(x^{4}+x^{2}+1)} + x\sqrt{x^{2}-x+1}\geq (x^{2}+1)\sqrt{1+x^{2}}$
$\sqrt{x(x^{4}+x^{2}+1)} + x\sqrt{x^{2}-x+1}\geq (x^{2}+1)\sqrt{1+x^{2}}$
Started By nightshade, 30-07-2012 - 10:08
#1
Posted 30-07-2012 - 10:08
- donghaidhtt likes this
#2
Posted 30-07-2012 - 12:56
Ngược dấu rồi bạn ơi!$\sqrt{x(x^{4}+x^{2}+1)} + x\sqrt{x^{2}-x+1}\geq (x^{2}+1)\sqrt{1+x^{2}}$
Ta có:
$VT=\sqrt{x^5+x^3+x}+\sqrt{x^4-x^3+x^2}$
$\leq \sqrt{2x^4+2x^4+2x^2+2x}$
Mặt khác ta có:
$VP=\sqrt{x^6+3x^4+3x^2+1}$
Áp dụng AM-GM ta có:
$x^6+x^4\geq 2x^5, x^2+1\geq 2x$
Do đó:
$VT\leq VP$
Nhưng dấu bằng không xảy ra nên: VT<VP
Edited by minh29995, 30-07-2012 - 12:56.
- hoangtrong2305, Mai Duc Khai and nightshade like this
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