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Inegalitate cu variabile mai mici decat 1/2 si respectiv 1

Posted: Wed Apr 08, 2009 11:16 am
by Claudiu Mindrila
Fie numerele reale \( a,\ b,\ c,\ d\in\left[0,\ 1\right] \) si \( x,\ y,\ z,\ t\in\left[0,\ \frac{1}{2}\right] \) astfel incat \( a+b+c+d=x+y+z+t=1 \). Sa se arate ca:

a) \( ax+by+cz+dt\ge\min\left\{ \frac{a+b}{2},\ \frac{b+c}{2},\ \frac{c+d}{2},\ \frac{d+a}{2},\ \frac{a+c}{2},\ \frac{b+d}{2}\right\} \ ; \)
b) \( ax+by+cz+dt\ge 54abcd. \)

Octavian Purcaru, O.N.M. 1996

Posted: Sun May 10, 2009 7:30 pm
by alex2008
\( WLOG\ a\le b\le c\le d \) . Deci minimul este \( \frac{a+b}{2} \) .

Avem ca :

\(
ax+by+cz+dt-\frac{a+b}{2}=y(b-a)+z(c-a)+t(d-a)-\frac{b-a}{2}\ge y(b-a)+z(b-a)+t(b-a)-\frac{b-a}{2}=(b-a)\left(\frac{1}{2}-x\right)\ge 0 \)