△ABC的三边a，b，c的倒数成等差数列，求证B＜π2

证明：方法一：已知1a+1c＝2b．得b＝2aca+c，a2+c2−b2＝a2+c2−(2aca+c)2≥2ac−4a2c2(a+c)2=2ac(1−2ac(a+c)2)≥2ac(1−2ac4ac)＞0．即cosB=a2+c2−b22ac＞0故B＜π2法2：反证法：假设B≥π2．则有b＞a＞0，b＞c＞...