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( x+y+z ) / 3 >= ( xyz ) 的立方根

( x+y+z )^3 / 27 >= ( xyz )

( xyz )^3 /27 >= ( xyz )

( xyz )^2 >= 27

xyz >= 根號 27

### Re: [高中]兩題數學.. 求XYZ..

lskuo 寫到:Continue (因為預覽時總是有些錯誤, 所以分段寫)

On the other hand, if y is known, then

x > =  (8-y)/(10-y) = 1 + 2/(y-10) > 1

So y > 10

Therefore, the only possible solution of y is 11.
If y = 11, then
10x + 11 = x^11 + 11^x + 11x
11 - x = x^11 + 11^x > x^11 + 11^1 = x^11 + 11
-x > x^11.  NO SUCH x (positive integer).

So the only solution is (x,y) = (1,9).

Because y(x-1) > 0,
10x - y(x-1) = x^y + y^x < 10x

x^y < 10x - x^y < = 10x
y ln x < ln 10 + ln x

(y-1) ln x < ln 10

ln x < ln 10 / (y-1)

2 < = x < 10^(1/(y-1))

log10(2)=0.3010 < 1/(y-1)
y < 1 + 1/log10(2) = 4.3

So now we know
2 < = y < 4.3, and
2 < = x < 10^(1/(y-1))

Try and error:
1. if y = 2, then x < 10
10x + 2 = x^2 + 2^x + 2x
x^2 = 8x + 2 - 2^x = even number
NO positive-integer solutions (try x = 2,4,6,8)

2. if y = 3, then 2 < = x < 10^(1/2) < 3.17. Possible x are 2 and 3.
10x + 3 - 3x = x^3 + 3^x
x = 2

3. if y = 4, then x < 10^(1/3) < 2.16. So the only possible x is 2.
10x + 4 - 4x = 16
x^y + y^x = 2^4 + 4^2 = 32
so (x,y) = (2,4) is not the solution.

The complete solutions are (x,y) = (1,9) and (2,3).
(x,y)=(2,3)也是一組解

### Re: [高中]兩題數學.. 求XYZ..

Continue (因為預覽時總是有些錯誤, 所以分段寫)

On the other hand, if y is known, then

x > =  (8-y)/(10-y) = 1 + 2/(y-10) > 1

So y > 10

Therefore, the only possible solution of y is 11.
If y = 11, then
10x + 11 = x^11 + 11^x + 11x
11 - x = x^11 + 11^x > x^11 + 11^1 = x^11 + 11
-x > x^11.  NO SUCH x (positive integer).

So the only solution is (x,y) = (1,9).

### Re: [高中]兩題數學.. 求XYZ..

Case 1: If x = 1, then 10 + y = 1 + y + y, so y =9.

Case 2: If y = 1, then 10x + 1 = x + 1 + x, so x = 0. Not positive integer!!!

Next step is to find solutions other than 1 included.
Note that if there is any other positive-integer solutions, both x and y would be greater than 1.

Rewrite the equation as
10x + y - xy = 10x - y(x-1) = x^y + y^x

Note that the minimum of x^y + y^x would be 2^2 + 2^2 = 8. So

10x - y(x-1) = x^y + y^x > =  8

If x is known, then one can find the upper bound of y.

y < =  (10x-8)/(x-1) = 10 + 2/(x-1) < =  12

So 1 < y < =  12

Note that y = 12 occurs when x=2. But if (x,y) = (2,12)
10x - y(x-1) = 20 - 12  = 8
x^y + y^x =  2^12 + 12^2 = 4096 + 144

Therefore (x,y) = (2,12) is NOT the solution.

xy-x-y+1=2，(x-1)(y-1)=2，
x-1=2，y-1=1，x=3，y=2

xyz>=xz^2 >= 4x
x+y+z<=3x 無解。