已知a,b,c∈R+ ,且abc=1 求证：a^2/(b+c)+b^2/(a+c)+c^2/(a+b)≥3/2
问题描述：
已知a,b,c∈R+ ,且abc=1 求证：a^2/(b+c)+b^2/(a+c)+c^2/(a+b)≥3/2
答
a,b,c∈R+均值定理a^2/(b+c)+(b+c)/4>=2√[a^2/(b+c)*(b+c)/4]=a同理b^2/(a+c)+(a+c)/4>=2√[b^2/(a+c)*(a+c)/4]=bc^2/(a+b)+(a+b)/4>=2√[c^2/(a+b)*(a+b)/4]=c即a^2/(b+c)+(b+c)/4+b^2/(a+c)+(a+c)/4+c^2/(a+b...