x>0,y>0,且x+y=15,則(x^2+4)^1/2+(y^2+16)^1/2之最小值最接近哪一個整數?

√(x² + 4) + √(y² + 16)
= √((x - 0)² + (0 - 2)²) + √((x - 15)² + (0 - 4)²)
= (0,2) 距 (x,0) + (x,0) 距 (15,4)

作(0,2)關於 x 軸之對稱點 (0,-2) ,
則 (0,2) 距 (x,0) + (x,0) 距 (15,4) = (0,-2) 距 (x,0) + (x,0) 距 (15,4) ≥ (0,-2) 距 (15,4) = √261,
直線 (0,-2) (15,4) 之方程為 (y+2)/x = (4+2)/15 ⇒ 2x = 5y + 10 與 x 軸 y = 0 交於 (5 , 0) ,
故 x = 5 > 0 , y = 10 > 0 時 √(x² + 4) + √(y² + 16) 之最小值 = √261 = 16.155... 最接近整數16。

Let a=x^2+4, b=y^2+16=(15-x)^2+16. a^(1/2)+b^(1/2) is the sum of two positive reals. By "arithmetic mean >= geometric mean", a^(1/2)+b^(1/2)>=[ab]^(1/2) and the minimum value occurs at a=b.
Thus solving for a=b => x=7.9 , and the minimum value is 2[(7.9)^2+4]^(1/2)]=16.298... Ans. integer 16.