MathDB
Inequality

Source: Romania 1996

January 27, 2010
inequalitiesinequalities unsolved

Problem Statement

a,b,c,d[0,1] a,b,c,d \in [0,1] and x,y,z,t[0,12] x,y,z,t \in [0, \frac{1}{2}] and a+b+c+d=x+y+z+t=1 a+b+c+d=x+y+z+t=1.prove that: (i) (i) ax+by+cz+dt ax+by+cz+dt \geq min(a+b2,b+c2,c+d2,d+a2,a+c2,b+d2) min( {\frac{a+b}{2} , \frac{b+c}{2} , \frac{c+d}{2} , \frac{d+a}{2} , \frac{a+c}{2} , \frac{b+d}{2} )} (ii) (ii) ax+by+cz+dt ax+by+cz+dt \geq 54abcd 54abcd