The problem of inequality is the problem of an inequality sinx siny

Because a + b a+b

So a + b [a+b (a + b)]a+b [a+b (a + b)].

i.e. ( a + b )[1 + a + b ] a + b )[1 + a + b ].

Divide both sides by the number greater than 0 [1+ a+b ]*1+ a + b ] to get the inequality.

Because |a|+|b|a+b So|a|*ㄧa+bㄧ+ㄧbㄧ*ㄧa+bㄧ+|a|+|b|≥|a|* a+b + b * a+b + a+b is ( a + b ) * 1+ a+b ) a+b * (1+ a + b ) so a + b 1+ a + b a+b 1+ a+b

A + B 1+ A + B A+B 1+ A+B That's what it means.

a|+|b|)/(1+|a|+|b|)>=|a+b|/(1+|a+b|)

Left=1 ((1 |.)a|+|b|)+1)

right=1 ((1 |.)a+b)+1)

a|+|b|>=|a+b|

1/(|a|+|b|)<=1/|a|b|

1/|a|+|b|)+1<=1/((1/|a+b)+1)1/((1/|a|+|b|)+1)>=1/(1/((1/|a+b)+1))

aㄧ+ㄧbㄧ/(1+ㄧaㄧ+ㄧbㄧ)≥a+bㄧ/(1+ㄧa+b|)

sinx-siny=2cos(x+y) 2sin(x-y) 2Make two radii in a circle with radius r so that the angle is , because the area of the triangle where the two radii are located is less than the area of the fan, so.

1/2r^2sinθ＜1/2θr^2

So sin

Thus |sinx-siny|=2|cos(x+y)/2||sin(x-y)/2|

2|sin(x-y)/2|

2|(x-y)/2|

x-y|

(1) From the question, xy

z=51，②4x8y

5z=300, multiply the formula left and right by 5 and subtract the formula left and right respectively, 5x-4x5y-8y=51*5-300, that is, y=x 315, multiply the formula left and right by 8 and subtract the formula left and right respectively, 8x-4x8z-5z=51*8-300, that is, z=-4x 336, from the question, 0 x 51,0 y 51,0 z 51, then 0 x 315 51,0 -4x 3

36 51, solution, 0 x 27

So there is, y=x 3

15，0≤x≤27；z=-4x/3

36，0≤x≤27；

2) From the question, substitute y and z obtained in (1) to obtain, 36) = p, simplified to obtain, 405 = p, then there is, 360

405 370, solution, 14 x 18, then x can take 14, 15, 16, 17, 18

So, when x = 14, y = x 3

15=59 3 (hectares), which does not match the title x, y, z are integers;

When x = 15, y = x 3

15 = 20 (ha), z = -4 3x

36=16 (ha);

When x = 16, y = x 3

15=61 3 (hectares), which does not match the title x, y, z are integers;

When x = 17, y = x 3

15=62 3 (hectares), which does not match the title x, y, and z are all integers;

When x = 18, y = x 3

15 = 21 (ha), z = -4 3x

36=12 (ha);

To sum up, this farm arranges rice, vegetables, and cotton planting areas respectively: 15 hectares, 20 hectares, and 16 hectares; or 18 hectares, 21 hectares, 12 hectares.

Set up the boss to enter x pieces late.

Option 1: 10*100 + x - 10)*100*= 85x +150

Scheme 2 price: 100 * * x = 89 x.

Hypothesis: Option 2 is better than option 1.

85x +150 > 89 x

The solution is x <150 4, because the missing branch x is an integer x <=37, and when scheme 1 is better than scheme 2, the solution is x >=38, so when the number of entries is greater than or equal to 38, scheme 1 is used.

If the number is less than or equal to 37, option 2 shall be adopted.

Solution: Add x grams of salt.

40+x)/(1000+x)>=10%

x>=200/3

A: Add at least 200 3 grams of table salt to make it a concentration of not less than 10%.

There are 40 grams of salt in the brine.

Then there is water 1000-40 = 960 grams.

To make it reach a concentration of 10%.

Then the water should not exceed 90%.

Later the total amount of brine should be reached.

960 90% = grams.

Add at least grams of table salt.

10(8/x+3/y)

2x+3y)(8/x+3/y)

16+6x/y+24y/x+9

25+6(x/y+4y/x)

x>0,y>0

x y+4y x>=2 (x y*4y x)=4, then x y=4y x is taken as an equal sign.

x=2y2x+3y=10

x=4,y=2

At this time, 10 (8 x + 3 y) > = 25 + 6 * 4 = 49 so x = 4, y = 2, the original minimum = 49 10

This problem seems to be a wrong condition, if there is only a, b, c>0 then the original formula can be reduced to.

In this form, the original form will never be greater than or equal to 6.

For a random variable x with a mean of 3 and a variance of 4, the lower bound of p(-2 < x < 8) is determined using Chebyshev's inequality.

2 .The mean is 3, the variance is 4, and the probability distribution of x is symmetrical with the mean 3, and the upper bound of p(x <= 0) is determined by Chebyshev's inequality.

The problem with this problem is that if a=b=c=1, then the original formula = 2+2+2=6<8

Hello, this (c a) b (a b) c (b c) a=c b+b c+a b+b a+a c+c a, the above equation can be greater than or equal to 2+2+2=6 by using the mean inequality

The first one is greater than or equal to 2

The second one is greater than or equal to 4

Greater than or equal to 2 and greater than or equal to 4

|x-1|+|2x-1|+|3x-1|+…2010x-1|+|2011x-1|minimum value.

When x=0, the original formula = 1+1+.1=2011When x=1 2011, the original formula = 1-1 2011+1-2 2011+.1-2011/2011

This is the minimum.

When x is about 1420, when the coefficient of x is 0, the inequality value is the smallest. I can't find a specific value though, around when x is around 1420. You can give it a try.