含根式不等式因技巧性较强,历年来颇受命题者喜爱,下面请欣赏几例. 一、三角代换例1 已知xi≥0,x0=0,sum from i=0 to nxi=1.求证:sum from i=1 to n xi/(1+x0+…+xi-1)(xi+…+xn)~(1/2)<π/2.证明 令x0+…+xi-1=sinθi-1,0=θ0≤θ1≤θ2≤…≤θn=π/2.则原式=sum from i=1 to n sinθi-sinθi-1/cosθi-1 The root-containing inequalities are very skillful and have been favored by propositioners over the years. Please appreciate a few examples below. 1. Triangle substitution example 1 It is known that xi≥0, x0=0, sum from i=0 to nxi=1. Proof: sum from i=1 to n xi/(1+x0+...+xi-1)(xi+...+xn)~(1/2)<π/2. Prove that x0+...+xi-1=sinθi-1 , 0 = θ0 ≤ θ1 ≤ θ2 ≤ ... ≤ θn = π/2. Then the original formula = sum from i = 1 to n sin θi - sin θi - 1 / cos θi - 1