求证：sqrt(a²+ab+b²)+sqrt(b²+bc+c²)≥a+b+c.已知a≥0,b≥0,c≥0.

Question

求证：sqrt(a²+ab+b²)+sqrt(b²+bc+c²)≥a+b+c.
已知a≥0,b≥0,c≥0.

Answer 1

sqrt(a²+ab+b²)=sqrt[(a+(b/2))^2+(3b^/4)]>=a+(b/2)
sqrt(b²+bc+c²)=sqrt[((b/2)+c)^2+(3b^/4)]>=(b/2)+c
所以
sqrt(a²+ab+b²)+sqrt(b²+bc+c²)≥a+(b/2)+(b/2)+c=a+b+c.