a>0,b>0,c>0,求证a^2/(b+c)+b^2/(a+c)+c^2/(a+b)≥(a+b+c)/2

a>0,b>0,c>0,求证a^2/(b+c)+b^2/(a+c)+c^2/(a+b)≥(a+b+c)/2
a>0,b>0,c>0,求证a^2/(b+c)+b^2/(a+c)+c^2/(a+b)≥(a+b+c)/2

a>0,b>0,c>0,求证a^2/(b+c)+b^2/(a+c)+c^2/(a+b)≥(a+b+c)/2
a>0,b>0,c>0
由柯西不等式得[(a/根号（b+c))^2+(b/根号（a+c))^2+(c/根号(a+b))^2][(根号（b+c))^2+(根号(a+c))^2+(根号（a+b))^2]≥[a/根号（b+c)*根号(b+c)+b/根号（a+c)*根号(a+c)+c/根号(a+b)*根号(a+b)]^2
即[a^2/(b+c)+b^2/(a+c)+c^2/a+b][(b+c)+(a+c)+(a+b)]≥(a+b+c)^2
因为(a+b+c)>0
所以a^2/(b+c)+b^2/(a+c)+c^2/(a+b)>=(a+b+c)/2