△ABC中.a+b=10.cosC=-1/2,求△周长的最小值

Question

△ABC中.a+b=10.cosC=-1/2,求△周长的最小值

Answer 1

根据余弦定理,c^2=a^2+b^2-2abcosC=(a+b)^2-ab=100-ab
a^2+b^2>=2ab=>(a+b)^2>=4ab
=>ab<=100/4=25
=>c^2=100-ab>=75
a+b+c>=10+根号75