Question

已知a,b,c∈R+且a²+b²+c²=1，求证a/1-a²+b/1-b²+c/1-c²≥3√3/2

Answer 1

证明(用局部不等式):
原式等价于a^2/a(1-a^2)+b^2/b(1-b^2)+c^2/c(1-c^2)≥(3V3)/2.
x(1-x^2)≤2/(3V3),则上式成立.
事实上,
x(1-x^2)
=V[2x^2(1-x^2)(1-x^2)/2]
≤V[1/2*((2x^2+ 1-x^2 + 1-x^2)/3)^3]
=V[1/2 * (2/3)^3]
=V (2^2/3^3)
=2/(3V3)
故不等式成立.