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题 设△ABC的三边BC、CA、AB长分别为a、b、c,P是△ABC所在平面上的任一点 ,P到顶点A、B、C的距离分别为R1、R2 、R3,证明或否定 :aR2 +R3+bR3+R1+cR1+R2≤ 32 (tan A2 +tan B2 +tan C2 )。(注 命题人对第一位解答正确者提供奖金 1 0 0元 )有奖解题擂台(68)$江西省南昌市华东
The three sides BC, CA, and AB of the term △ABC are a, b, and c, and P is any point on the plane where △ABC is located. The distance from P to the vertices A, B, and C is R1, R2, and R3, respectively. Proof or negation: aR2 + R3 + bR3 + R1 + cR1 + R2 ≤ 32 (tan A2 + tan B2 + tan C2 ). (Note: The proposition person provides the bonus for the first person who answers the correct 100 yuan.) There is the prize solution for the encyclopedia (68) $ East China, Nanchang, Jiangxi Province.