Inegalitate

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heman
Euclid
Posts: 39
Joined: Fri Sep 28, 2007 7:36 pm

Inegalitate

Post by heman »

Pentru x,y,z \( \in \) \( [\frac{1} {3},\frac{2} {3}] \) are loc inegalitatea:
1 \( \ge \) \( \sqrt[3] {xyz} \)+\( \frac {2} {3(x+y+z)} \)
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Radu Titiu
Thales
Posts: 155
Joined: Fri Sep 28, 2007 5:05 pm
Location: Mures \Bucuresti

Post by Radu Titiu »

Suficient sa aratam ca
\( 1-\frac{S}{3}-\frac{2}{3S}\geq 0 \), unde S=x+y+z.
Echivalent:
\( S^2-3S+2\leq 0 \Leftrightarrow \) \( S\in [1,2] \), care se obtine folosind ipoteza.
A mathematician is a machine for turning coffee into theorems.
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