Cho $x^{2} + y^{2} + z^{2} = 1$. Chứng minh rằng :
$xyz + 2\left ( 1 + x + y + z +xy + yz + zx \right ) \geqslant 0$.
$xyz + 2\left ( 1 + x + y + z +xy + yz + zx \right ) \geqslant 0$
Started By tramyvodoi, 26-01-2013 - 15:59
#1
Posted 26-01-2013 - 15:59
tramyvodoi
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Thượng úy
- haisupham, Anh la ai, Nguyen Minh Tuan B and 2 others like this
#2
Posted 30-04-2013 - 10:37
Tienanh tx
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$\Omega \textbf{Bùi Tiến Anh} \Omega$
$\oplus$ Theo giả thiết suy ra $ - 1 \leq a,b,c \leq 1$
$\Longrightarrow$ $(1+a)(1+b)(1+c) \geq 0 \Longrightarrow 1+a+b+c+ab+bc+ca + abc \geq 0 (1)$
$\oplus$ Ta có:
$1+ a+ b+c+ ab+bc+ca = \dfrac{1}{2}[1+2ab+2ac+2ca + 2(a+b+c) + 1]= \frac{1}{2}[(a+b+c)^2+2(a+b+c) + 1] = \frac{1}{2}(a+b+c+1)^2 \geq 0 (2)$
Cộng (1) và (2) suy ra đpcm
- caybutbixanh likes this
$\cdot$ $( - 1) = {( - 1)^5} = {( - 1)^{2.\frac{5}{2}}} = {\left[ {{{( - 1)}^2}} \right]^{\frac{5}{2}}} = {1^{\frac{5}{2}}} =\sqrt{1}= 1$
$\cdot$ $\dfrac{0}{0}=\dfrac{100-100}{100-100}=\dfrac{10.10-10.10}{10.10-10.10}=\dfrac{10^2-10^2}{10(10-10)}=\dfrac{(10-10)(10+10)}{10(10-10)}=\dfrac{20}{10}=2$
$\cdot$ $\pi\approx 2^{5^{0,4}}-0,6-\left(\frac{0,3^{9}}{7}\right)^{0,8^{0,1}}$
$\cdot$ $ - 2 = \sqrt[3]{{ - 8}} = {( - 8)^{\frac{1}{3}}} = {( - 8)^{\frac{2}{6}}} = {\left[ {{{( - 8)}^2}} \right]^{\frac{1}{6}}} = {64^{\frac{1}{6}}} = \sqrt[6]{{64}} = 2$
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