4 inequality questions, high reward for the highest

1.Solution: Because (x 2-y 2)(x+y) = (x+y) 2(x-y) compared to (x 2+y 2)(x-y).

1* When x>y Since x is less than 0 and y is greater than 0, we get: (x+y) 2>(x 2+y 2).

Divide by (x-y) so (x 2-y 2) (x+y) > (x 2+y 2) (x-y).

2* When x(x 2+y 2).

Divide by (x-y) so (x 2-y 2) (x+y) > (x 2+y 2) (x-y).

2.Proof that 4 (a-3) +a is greater than or equal to 7 is equivalent to.

4+a(a-3) is greater than or equal to 7(a-3) is equivalent to.

A 2-10a+25 is greater than or equal to 0

A-5) 2 is greater than or equal to 0.

Proven. 3.Solution: Because xy=1.

x^2+y^2)/x-y=[(x-y)^2+2]/x-y

x-y)+[2/(x-y)]

2 followed by 2 (x-y>0 because x>>y0).

If and only if (x-y)=[2 (x-y)].

The solution is x = 2 + 6 divided by 2

y = with 6 - with 2 divided by 4

4.Solution: Compare differences.

loga(t+1/2)]-1/2logat]

loga (with t+1 (2 with t)

Because with t+1 (2 with t)>=gen2>1

When a>1 loga(t+1 2)]-1 2logat]>0

i.e. loga(t+1 2)]>1 2logat].

When 0 is loga(t+1 2)]<1 2logat].

Mainly the third question, I only did it roughly, I didn't make a draft, please choose the answer yourself.

1.If you know that x is less than 0 and y is greater than 0, try to compare the size of (x square + y squared) (x-y) and (x square - y squared) (x+y).

x square + y square) (x-y) - (x square - y squared) (x + y).

x^3+xy^2-x^2y-y^3-x^3-x^2y+xy^2+y^3

2x^2y<0

x square + y square) (x-y) - (x square - y squared) (x + y).

2.If we know that a is greater than 3, verify that 4 (a-3) +a is greater than or equal to 7

4/（a-3） +a-7=4/(a-3)+(a-3)-4>=2√[4(a-3)/(a-3)]-4=0

When 4 (a-3) = a-3, i.e. a = 5, the equal sign holds.

4/（a-3） +a >=7

3.It is known that x is greater than y and greater than 0, and xy=1 is used to find the minimum value of (x square + y squared) (x-y) and the value of x,y at this time.

Let x=tana,,45 degrees = 2 2

sina) 2-(cosa) 2] (sinacosa)=2(sinacosa) [(sina) 2-(cosa) 2].

The minimum value is 2 2

cos2a)^2=(sin2a)^2/2

tan2a)^2=1/2

2a = 135 degrees.

tan135=2tana/[1-(tana)^2]=-√2/2

2x/(1-x^2)=-√2/2

2x^2-4x-√2=0

x=(4+2√6)/(2√2)=√2+√3

y=1/x=√3-√2

Let a be greater than 0, a is not equal to 1, and t is greater than 0, try to compare the size of 1 2logat and loga (t+1 2).

1/2logat=loga√t

t+1/2)^2=t^2+t+1/4>t

t+1/2)>√t

When 0loga(t+1 2).

When a>1.

1/2logat

1.Because: x<0,y>0, so 2xy<0,x 2+2xy+y 2 < x 2+y 2

x 2+y 2)*(x-y)<(x+y) 2 *(x-y) then: (x 2+y 2)*(x-y)<(x 2-y 2)*(x+y)2Because a>3, a-3>0

4 (a-3) +4 + (a-3) = 2>0.

3.(x 2+y 2) (x-y)=(x-y)+2 (x-y)>=2, where x=(root number 6 -2) 2, y=(root number 6 +2) 2

4.Because t>0, t+1 2> root number t is 0loga (t+1 2).

When a>1, loga(root t) = 1 2logat

1。Do the bad method.

x 2+y 2)(x-y)-(x 2-y 2)(x+y)=(x 3-x 2y+xy 2-y 3)-(x 3+x 2y-xy 2-y 3)=-2x 2y+2xy 2=2xy(y-x)<0 is negative.

3 a-3>0

Then 4 (a-3)+a=4 (a-3)+(a-3)+3>=2 times the root number (4 (a-3)*(a-3) +3>=7

This problem uses the mean inequality.

3. x=1/y>y>0

1 y-y>0 (1-y 2) y>0 00 discriminant <0) if 01 the latter big ok pull.

0>2xy

x^2+y^2>x^2+y^2+2xy

x^2+y^2)*(x-y)<(x^2+y^2+2xy)(x-y)=(x+y)^2(x-y)=(x^2-y^2)(x+y)

a-5)^2>0

a^2-10a+25>0

a^2-3a+4>7a-21

4+a(a-3)>7(a-3)

4/(a-3)+a>7

t=x-y>0

x^2+y^2)/(x-y)

x^2+y^2-2xy)/(x-y)+2xy/(x-y)x-y+2/(x-y)

t+2 t=2*root number 2

There seems to be a problem.

1: To make a difference comparison, the first term minus the second term (first) gets 2xy(y-x) and y is greater than 0 and greater than x, so this item is greater than zero.

2. Prov: 4 (a 3)+a=[4 (a 3)+a 3]+3 is greater than or equal to 2 2+3=7, where the arithmetic geometric mean inequality is used in the middle brackets.

Three: Bring the reciprocal of y to x into the desire formula, find the point where the derivative is equal to zero after derivation, simplify x 6 - 3 x 4 - 3x 2 + 1 = 0, and further factor it to get (x 2 + 1) (x 4 - 4x 2 + 1) = 0 Obviously, x square plus one is greater than zero, as long as the latter term is solved, the ......... is obtainedI've figured it out, but sorry I won't be able to do it.

Four: I guess you are confused by the size A, in short, it is good to discuss it in a situation.

ax-1)(x-1)<0

If 1 a>1, that is, 01, 1 a=2 (4x*9y)+2xy=12 s+2s (take 4x=9y).

2s+12 s-320=2( s+16)( s-10)<=0 s<=10, s<=100, that is, the maximum value of s is 100s=xy=100, and 4x=9y

x=15, y=20 3, that is, the iron grid is 15m long

3) ax 2+bx+c<0 solution set is empty set silver wheel early, a>0, =b 2-4ac<0

ab>0,a>0,b>0

a 2+b 2-2b=a 2+(b-1) 2-1a 2>0,b-1) 2> finch = 0 (take equal b=1) a 2+b 2-2b>-1, i.e., the range is (-1, infinite).

Solution: (1) The number of eligible people is

2) The number of eligible people is: 108.

45x+30（8-x）》=330

x》=6, so x=6 or 7 or 8

Cost = 400x + 280 (8-x).

120x+2240

When x=6 there is a minimum value of 2960

Because a>-2, b>0 and a+b=8 are merged with b less than 10 according to the mean inequality, < (a+2)+b> 2 is greater than or equal to how much (answer) because the value is the maximum when a+2=b, so a=3, b=5 is the maximum, so 10 2 is equal to 5

Because the hall is an even function, the left and right are equal, and it is a subtractive function, so at (0,+ edge x>a is a small absolute before 0, and according to symmetry, x<-a is less than 0.

(3x2-14x+14-x2+6x-8) x2-6x+8>02(x-1)(x-3) (x-4)(x-2)>0 inequality is equivalent to (x-1)(x-2)(x-3)(x-4)>0 and x (-1) (2,3) (4,+.)

Use the thread-and-needle method to do this kind of problem.

Multiply the denominator and it becomes a 1 x 2 unknown quadratic inequality problem.

The maximum value is negative.

That is, the image of this quadratic function is open downward.

And it has no intersection with the x-axis, that is, the discriminant formula is less than 0

So it's available. a<0△=16-4a(a-3)

4a²+12a+16

4(a²-3a-4)

4(a-4)(a+1)<0

Get A<-1 or A>4

To sum up a<-1

a<0 and a<-1 intersect to get a<-1a<0 and a>4 to intersect to get an empty set, so the range is a<-1