若D在AB線段上，E在AC線段上且三角形ADE之面積為三角形ABC面積的1/3，求DE線段長度最小值為何?

令 a = BC , b = AC , c = AB
設 AD = s*AB , AE = t*AC

ΔADE面積 / ΔABC面積 = AD*AE / ( AB*AC ) = st = 1/3
3st = 1
st = 1/3 且 t = 1/( 3s )

DE²
= AD² + AE² - 2*AD*AE*cos∠A , 餘弦定理
= s²(AB²) + t²(AC²) - 2st*AB*AC*cos∠A
= s²c² + t²b² - 2stcb*cos∠A
= c²s² + b²/(9s²) - (2/3)bc*cos∠A , 因為 st = 1/3 且 t = 1/( 3s )

[ (cs)² + ( b / (3s) )² ] * [ 1² + 1² ] ≧ [ cs + b/(3s) ]² , 柯西不等式
2 [ c²s² + b²/(9s²) ] ≧ c²s² + b²/(9s²) + (2/3)bc
c²s² + b²/(9s²) ≧ (2/3)bc ..... (1式)

柯西不等式該式中, 等號成立條件為 cs = b/(3s)
b = 3cs²
s = √[ b / (3c) ]
AD = s*AB = sc = √(bc/3) = √( AC * AB / 3 )
AE = t*AC = tb = b/(3s) = (b/3)√(3c/b) = √(bc/3) = √( AC * AB / 3 )

DE²
= c²s² + b²/(9s²) - (2/3)bc*cos∠A
≧ (2/3)bc - (2/3)bc*cos∠A , 由 (1式)
= (2/3)bc*( 1 - cos∠A )
= (2/3)AC*AB*( 1 - cos∠A )

DE ≧ √[ (2/3)AC*AB*( 1 - cos∠A ) ]

Ans:
當 AD = AE = √( AC * AB / 3 ) 時 ,
DE 有最小值 √[ (2/3)AC*AB*( 1 - cos∠A ) ]