The inequality of the upper period of the second year of high school, the inequality of the second y

1) a,b,c (0,+ then add 3 numbers at the same time and divide by 3, 3=3(abc+1) 3abc=1+1 abc

If a, b, and c are all integers, then 1+1 abc is less than 2, but all 3 numbers alone are greater than 2

If a, b, and c are all fractions, then 1+1 abc is greater than 2But all 3 numbers alone are greater than 2

If at least 1 of a, b, and c is a fraction, then 1+1 abc is greater than or less than 2But all 3 numbers alone are greater than 2

If at least 2 of a, b, and c are scores, then 1+1 abc is less than or greater than 2, but all 3 numbers alone are greater than 2 A should be selected for this question

2) x+y+z=81, find the maximum value, you can deduce that x=26 y=27 z=28 then the final result of this problem should be 19656, and 28*28*28=21952This question is misunderstood.

If x*y*z=81, find the maximum value, and you can deduce x= 1 y= 3 z=9Then the final result of this question should be 80. Still unanswered.

So the answer to this question may be d

3) Backwards, assuming a=b=c,, only b meets the condition a=1 3

, with counter-evidence.

2, d, with a fundamental inequality.

3, b, the same method as above.

4.I don't know if there are parentheses after that divide, so I didn't do it.

Since x>=1, the numerator and denominator are in the same hall circle, and the manuscript file is divided by x

f(x)=x/(x2+2(a+2)x+3a)1/[x+2(a+2)+3a/x]

1 [2 root number 3a+2 (a+2)].

If you can use the mean inequality, both are greater than 0

i.e. x>0 3a x>0

And you can get the equal sign, so you can get x = 3a

Since x>=1

So 3a>=1

So a> key volley = 1 3

It is known that a 0, b 0, a + 2 b = 1

So (2 a+b) = (2 a+b)(a+2 b) = 2+4 ab+ab+2 4+2 ((4 ab)*ab)=8

The minimum value taken if and only if 4 ab=ab, i.e., ab=2, i.e., a=1, 2, b=4.

Solution: From the problem and the "Cauchy inequality", it can be seen that (2 a)+b=[a+(2 b)] 2 a)+b] (2+ 2) =8

i.e. (2 a) + b 8The equal sign is obtained only when a=1, 2, b=4. ∴[2/a)+b]min=8.

Because: a>0, b>0 a+2 b=1

2/a)+b]*[a+(2/b)]=2+ab+(4/ab)+2=ab+(4/ab)+4

ab+(4 ab)+4 is greater than or equal to 2 root number [ab*(4 ab)]=4, so: ab+(4 ab) is greater than or equal to 4-4=0(2 a) + b is greater than or equal to 0

The minimum value of 2 a) +b is 0

In fact, it is monotonically decreasing.

So a b has to prove to be monotonically decreasing.

Immediate (n+1)- n n- (n-1)Immediate (n+1)+ n-1) 2 n

i.e. ( (n+1)+ n-1)) 2 (2 n) 2 i.e. n+1+n-1+2 (n+1) (n-1) 4n i.e. (n2-1) n

That is, n 2-1 n 2 is proven!

Solution: 3(1+x 2+x 4)-(1+x+x 2) 2=3+3x 2+3x 4-(1+x+x 2+x+x 2+x 3+x 2+x 3+x 3+x 4).

2+2x^4-2x-2x^3

2(x^4-x^3-x+1)

2(x-1)(x^3-1)

2(x-1)^2*(1+x+x^2)

Because 1+x+x 2=(x+1 2) 2+3 4 3 4 0 and x≠1 (x-1) 2 0

So 2(x-1) 2*(1+x+x 2) 0 so 3(1+x 2+x 4) (1+x+x 2) 2

1. Assuming a >=b, just prove f(a)-f(b)<=| a-b|, and assume a>b, only 1+a - 1+b <=a-b, i.e., a- 1+a >=b- 1+b (*true;

Construct y=x- 1+x, which proves that y is a subtractive function; So (*) is true; So when a >=b and a>b, the proposition is true;

a=b and a>b, the proposition is true;

2, in the same way, a can be obtained, so when a is not equal to b, |f(a)-f(b)|<=|a-b|Total establishment.

Note: This question focuses on logical analysis ability.

1<=a+b<=5===>1/2<=(a+b)/2<=5/2---1)

1<=a-b<=3==>-5/2<=5(a-b)/2<=15/2---2)

1) +2) Got:

1/2-5/2<=3a-2b<=5/2+15/2-2<=3a-2b<=10

Why can't it be counted that way? Do you need linear programming? How simple and how to come)