How many inequality questions are asked, how do you do this basic inequality problem?

p=e^x+e^-x>=2

q=(sinx+cosx)^2=1+sin2x<=2p>=q

x>0,y>0

then x+y>=2(xy) (1 2).

xy-(x+y)<=xy-2(xy) (1 2) then xy-2(xy) (1 2)>=1

xy-2(xy)^(1/2)-1>=0

The solution is (xy) (1 2)<=1-2 (1 2), (xy) (1 2)>=1+2 (1 2).

xy>0

xy>=(1+2^(1/2))^2=3+2*2^(1/2)xy-(x+y)<=(x+y)^2/4-(x+y)x+y)^2/4-(x+y)>=1

x+y)^2-4(x+y)-4>=0

x+y>=2(1+√2 )

The minimum value is 2 (1 + 2).

There was a mistake upstairs. First, the x in p q cannot be taken at the same time. Nor can you get the one in question 2 at the same time.

2p=e^x+e^-x>=2

But here x does not satisfy both conditions.

When x=0, the minimum value of 2 is taken, and the maximum value of q is not taken. Therefore, cxy<=(x+y) 2 2 is chosen

x+y)^2/2-(x+y)》1

Let t=(x+y) 》0 t 2-2t-2》0t》(2+ 12) 2

The minimum value is 1 + 3

As shown in the figure below, this question uses the commutation method.

<(lgx)^2

lgx(lgx-2)>0

lgx>2 or lgx<0

x>100 or 00

Discriminant union 0

The original inequality is constant for any a.

In summary-11 alpha}

1 is the solution of the left side of the inequality = 0 is 1 2, 2 is directly brought in, a = plus or minus 12 when x 2>1, m<(2x-1) (x 2-1) is constant, so (2x-1) (x 2-1) > 2, solution (1-root3) 21 or x<-1, then 1 circle or (2x-1) (x 2-1) is constant, so (2x-1) (x 2-1) <-2, solution x<(-1-root7) 2 or x>(-1+root7) 2

-11,lg(ab)>0 , so lgc "lg(ab)c"ab

1.x+2y=1, multiplied by x:

2xy + x^2 = x

xy = (x - x^2)/2

xy - 1/8 = (x - x^2)/2 - 1/8= -(4x^2 - 4x +1)/8

(2x -1)^2/8 ≤ 0

So xy 1 8

x = 1 2, -2x -1) 2 8 = 0, xy = 1 8

2.sqrt[x(1 - x)] sqrt: square root).

sqrt(x - x^2)

sqrt(-x 2 + x -1 4 + 1 4) = sqrt [-x - 1 2) 2 +1 4]When x = 1 2, (x - 1 2) 2 = 0, sqrt[x(1 - x)] maximum (=1 2).

Question 1: |x-4|-|x-3|It can be seen as the difference between the distance from a point on the coordinate axis to the point with coordinates 4 and 3, and in order for the inequality to be solved, a must be rounded up to be less than the maximum value of the distance difference.

Find its maximum value below; When x is between 4 and 3, it is clear that the maximum value is 1, i.e. x = 3.

When x is on the left side of 3 and x is on the right side of 4, they are 1 and -1, respectively

So a less than 1 makes the inequality a solution.

Question 2: Finding the range of y=1-2x-3 x is finding the range of (2x+3 x).

Combined with a 2 + b 2>2ab, when x > 0, 2x+3 x>=2 * root number 6

When x<0, 2x+3 x<=-2*root number 6

Then the range of y is less than or equal to 1-2 * root number 6 and greater than or equal to 1 + 2 * root number 6

Question 3: Transform 1 x+4 y=1 into x=y (y-4)=1+4 (y-4).

x+y=1+4/（y-4）+y=1+4/（y-4）+y-4+4>=1+4-4=1

So the minimum value of x+y is 1

Question 4: It's a bit troublesome, let me do the math.

1、b^2-4ac=（-a-c)^2-4ac=a^2-2ac+c^2=(a-c)^2>0

2、x^2/(x-1)>0

x>1 is set to x-1=t>0, then x=1+t

y=log2[(t^2+2t+1)/t]=log2(t+1/t+2)≥log2(2+2)=2

The value range is [2,+

3、a^a·b^b/(a^b·b^a)

a/b)^a·(b/a)^b

a/b)^(a-b)

1) When a>b>0, ab>1, a-b>0(ab) (a-b)>1

a^a·b^b>a^b·b^a

2) When a=b>0, a-b=1, a-b=0(a b) (a-b)=1

a^a·b^b=a^b·b^a

3) When 01a a·b b>a b·b a

So a a·b a b·b a

1. 9≤a＜12

3.1 b<1 a<0

4. x2,x1,x3

x1+x2=a1, ①

x2+x3=a2 ,x3+x1=a3，③

x1-x3=a1-a2 0, x1 x3- x2-x1=a2-a3 0,, x2 x1x2 x1 x3