Mathematics Times – July 2019

  36 4.4.9 0y^2   y^2 4 0

 y 2 or y 2 (extraneous because

whenx y 0, 0).

From y 2 ,we know the smallest value of y

is 2, which occurs when z^32 or x^12.

Method 2:

 

 

 

(^242) 2 1 9
4 8 13
6 1 6 1
x x x x
x x
    

 
 
 
2
4 1 9
6 1
x
x
 

.
Let
 
 
(^22)
4 1 9 4 9 2 3
1 ,
6 1 6 3 2
x b b
x b
x b b
  
    
.
Let
2 2 3 1
.
3 3 2
b b
a a
b a
   .
From AM-GM,we known that
1
a 2
a
  for
positive a.
Thus the smallest value of  
4 8 13^2
6 1
x x
x
 
 is 2,
which occurs when a 1 , or b^32 , or
1
x 2.
2.Sol: Let
1 x^21
x t
x x

  .
Since x t 1, 2.
2
2 2
1 8 1 8 8 8
2.
1 1
x x x
x t t
x x x x t t

      
 
(^) 4 2
If and only if
8
t
t
 , or t2 2, or
x 2 1, will the smallest value of
Method 2:
2.Sol:
3.Sol:
4.Sol:
5.Sol:
6.Sol:
4 2
be achieved.
3.Sol:y x x x x       (^2)  (^3)  (^1)  4 35
     x x x x^2 5 6^2 5 4 35
The average of x x^2  5 6and
x x^2  5 4 is x x^2  5 5
Let t x x  ^2 5 5.
We have y t t      1 1 35  t^234.
We see that the smallest value is 34 when
t 0.
That is, when
(^) x x^2   5 5 0 or min
5 5
, 34
2
x y
 
 .
4.Sol:Let
1
t x
x
 .
Theny t ^23.
When t0,or x y 1, min 3.
5.Sol:Let x y u  and xy v
So we have u v u^2   3 3 .
v u u  ^2 3 9 (1)
   
2 2 2 2 2
x y u v u u u       2 3 9 3
 27 (2)
Since x and y are two roots of equations
t ut v^2    0 , we have
  u v^2 4 0 (3)
Substituting (1) into (3) yields
u u u^2     4 ^2 3 9 0.
Solving this inequality, we get    2 u 6 .(4)
Therefore x y^2  ^2 max^27 when u 3 and
v u u   ^2 3 9 9.
6.Sol:x x y^2   2 4 5
(^2) 2 5
2
2
x x
y
 
 
(^2) 2 5 2 (^2) 2 5
2
2 2
x x x x x
x y x
    
   
   
1 2 1 2 9
4 5 2
2 2 2
     x x x
Since x 2 0^2 and  
1 2
2 0
2
  x , the