mateforum.ro

Determinati numerele reale \( a \), \( b \), \( c \), \( d \) astfel incat \( a+b+c+d=20 \) si \( ab+ac+ad+bc+bd+cd=150 \)

\( (a+b+c+d)^2=a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd \)

Deci \( \sum_{cyc} {a^2}=400-2\cdot150=100 \)

Insa \( (a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2=3\sum {a^2}-2(ab+ac+ad+bc+bd+cd)=300-300=0 \)

Asadar \( a=b=c=d=5 \)