algebra 代數

Consider
　　　(√b－√c)²＞＝0
＝＞　　b－2√(bc)＋c＞＝0
＝＞　　b＋c＞＝2√(bc)

Therefore, we have

b＋c＋1/(bc)＞＝2√(bc)＋1/(bc)

Let x＝bc, we have

b＋c＋1/(bc)＞＝2√x＋1/x

Let f(x)＝2√x＋1/x, we have

df(x)/dx＝2(1/2)/√x－1/x²＝1/√x－1/x²
d²f(x)/dx²＝－(1/2)/(√x)³＋2/x³　　　(1)

To find the critical value of f(x), we set

　　　df(x)/dx＝0
＝＞　1/√x－1/x²＝0
＝＞　1/√x＝1/x²
＝＞　x²/√x＝1　　　(b＞0,c＞0　＝＞　bc＞0　＝＞　x＞0)
＝＞　(√x)³＝1
＝＞　x＝1

Therefore, the critical value of f(x) occurs when x＝1.

From (1), we have when x＝1

d²f(x)/dx²＝－(1/2)/(√1)³＋2/1³＝3/2＞0

Therefore, f(1)＝2√1＋1/1＝3 is the minimum value of f(x).

Therefore, b＋c＋1/(bc)＞＝f(x)＞＝3

It is obvious that when b＝c＝1, we have

b＋c＋1/(bc)＝1＋1＋1/(1．1)＝3

Therefore, the minimum value of b＋c＋1/(bc) is 3.

f(x) = x + c + 1/(cx)
f`(x) = 1 - (1/c) (1/x^2)
f``(x) = (2/c) (1/x^3) > 0
set f`(x) = 0, x = 1/sqrt(c)
min f(x) = 1/sqrt(c) + c + (1/c) (sqrt(c))
= 2/sqrt(c) + c
------------------------------------------
g(x) = 2/sqrt(x) + x
g`(x) = -1/x^(3/2) + 1
g``(x) > 0
set g`(x) = 0, x = 1
min g(x) = 2/sqrt(1) + 1 = 3
so, 最小值 = 3