Question

不等式（2）设a,b,c＞0,求证a/(b+c)^2+b/(c+a)^2+c/(a+b)^2≥9/4(a+b+c)

Answer 1

[a/(b+c)^2+b/(c+a)^2+c/(a+b)^2 ]*(a+b+c)

= a/(b+c) + b/(c+a) + c/(a+b)+a^2/(b+c)^2 + b^2/(c+a)^2 + c^2/(a+b)^2

a/(b+c) + b/(c+a) + c/(a+b)

= (a+b+c)(1/(b+c) + 1/(c+a) + 1/(a+b)) -3

≥ 9/2 - 3

= 3/2

a^2/(b+c)^2 + b^2/(c+a)^2 + c^2/(a+b)^2

≥1/2[ a^2/(b^2+c^2) + b^2/(c^2+a^2) + c^2/(a^2+b^2)]

≥ 3/4

a/(b+c) + b/(c+a) + c/(a+b)+a^2/(b+c)^2 + b^2/(c+a)^2 + c^2/(a+b)^2 ≥ 9/4

a/(b+c)^2+b/(c+a)^2+c/(a+b)^2≥9/4(a+b+c)