Page 112 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

3.73 x,y,z are positive numbers, show (x+1) 3 + (y+1) 3 + (z+1) 3 ≥ 81 .
y
z
4
x
3
3
Proof: Since x,y,z > 0,Mean Inequality implies that (x+1) + 27 y + 27 ≥ 3 3 (x+1) 3 · 27 y · 27 = 27 (x + 1) ⇒ (x+1) ≥ 27 (x − y)+ 27
y 2 4 y 2 4 2 y 2 4

(x+1) 3 + 27 y + 27 ≥ 3 3 (x+1) 3 27 27 = 27 (x+1) 3 27 (x − y)+ 27 . Similarly, we can obtain (y+1) 3 27 27 (z+1) 3 27 27
y 2 4 y · 2 y · 4 2 (x + 1) ⇒ y ≥ 2 4 z ≥ 2 (y − z)+ 4 , x ≥ 2 (z − x)+ 4
(y+1) 3 27 (y − z)+ 27 (z+1) 3 27 (z − x)+ 27 . Add them up to obtain the aimed inequality.
,
z ≥ 2 4 x ≥ 2 4
√ √
3.74 m, n are positive numbers, show m +1 > n holds if and only if for any
√
x
x> 1, mx + x−1 > n .
360°
√ √
1
x
Proof: m, n > 0,x − 1 > 0, then mx+ x−1 = mx−m+m+ x−1+1 =[m(x−1)+ x−1 ]+m+1 ≥ 2 m+m+1 = ( m+1) 2

x−1
√ √ 1
2
mx+ x = mx−m+m+ x−1+1 =[m(x−1)+ 1 ]+m+1 ≥ 2 m+m+1 = ( m+1) . If and only if m(x − 1) = 1 , i.e. x = 1+ √ , mx + x has
x−1 x−1 x−1 x−1 m x−1
√ √ thinking.
x
2
the minimum value ( m + 1) . Hence, mx + x−1 > n for any x> 1 if and only if
√
( m + 1) >n , i.e. √ m +1 > √ n . 360°
2
thinking.

360°
thinking.

360°
thinking.

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