Let the midpoint of BC be d

AG = 2/3 * AB = 1/3(a b+AC)= AE/3λ+AF/3μ

According to the vector three-point * * * line theorem, the GEF three-point * * line1/3λ+1/3μ =1= >1/λ+1/μ = 3 is a constant value.

3 (λ+μ) = (1/λ+1/μ) (λ+μ) = 2+(λ/μ/λ) ≥ 2 √ (λ/μ/λ) is the substitution of 3, and then the minimum value is obtained by using the basic inequality.

Similarly, why can there be 0 <1/λ+1/μ≤ [(1/λ+1/μ)/2] The answer here may be problematic.

On the basic inequality A+B > The square of =2 ab is on both sides of ab < = [(a+b)/2] 2.

So1/λ *1/μ≤ [(1/λ+1/μ)/2] 2