Question

已知函数f(x)＝x²＋λx,p、q、r为△ABC的三边

Answer 1

答：λ>-3 .

分析：f(r)-f(q)>0

r²＋λr-(q²＋λq)=r²-q²＋λr-λq=(r+q)(r-q)+λ(r-q)

=(r-q)(r+q+λ)>0 ① 又q＜r,∴(r+q+λ)>0 ,λ>-(r+q)

同理, (q-p)(q+p+λ)>0 ② 又p＜q,∴(q+p+λ)>0 ,λ>-(p+q)

(r-p)(r+p+λ)>0 ③ 又p＜q,∴(r+p+λ)>0 ,λ>-(r+q)

又p＜q＜r,∴λ>最大的-(p+q),

p、q、r三者均为正整数,p＜q＜r,r min=1,qmin=2

λ>-3