Some Inequalities in High School! I would like to ask for your help!!

1.The maximum value can be obtained by using the boundedness of linear programming or the coordination method (x+2y=a(2x+y)+b(x+3y)).

2.It can be known that when a<0 there will be a set of solutions with left and right limits, and adding a negative sign to the original inequality becomes ", which can be solved by using Veda's theorem, and the answer is b

3.Because 00, and then the answer can be found using the one-step mean inequality, ie.

3x(8-3x) (3x+8-3x) 2 4=16, so the maximum value of f(x) is 4 a

4.First of all, you can exclude the a term, because the inequality of a=0 does not hold, and then take a=-1 to get x 2+2x+2>0 constant, so d is chosen. If it is a general calculation, it can be calculated as a<0, <0.

5.Constant to establish the problem, the mantra is greater than the big, less than the small. x-1>0, you can make t=x-1, t>0, and then use the mean inequality, the original inequality = t+(1 t)+1 2+1=3, so m 3, choose d.

bbadd is correct.

The first comrade. The problem has been clearly determined to be a quadratic function, so a must not be equal to 0

Yes 9. The process of finding is that the condition is set by the question, a> ridge 2, b>3, b 3=a (a-2)=1+2 (a-2).

2a+b/3=2a+1+2/(a-2)=2[(a-2)+1/(a-2)]+5。Application of the Fundamental Inequality, 2a+b 3 9. Where, a=3, b=9, "= holds.

Let u=2 a, v=3 b, then u,v>0, u+v=1, a=2 u, b=3 v, y=2a+b 3=4 u+1 v=4 u+1 (1-u).

4-3u) (u-u 2) with x=4 3-u, then u=4 3-x, y=3x [4 3-x-(4 3-x) 2].

3x/(-4/9+5x/3-x^2)

4 (9x)+x 4 3, when x=2 Tongsen 3, u=1 3, take the equal sign, so y 3 (5 3-4 guess mill 3) = 9, so the smallest value of y is 9

1/x+5x/y

3x+4y)/5x+5x/y

3/5+(4y/5x+5x/y)

3/5+2√(4y/5x▪5x/y)

Only if 4y 5x = 5x y, 4y = 25x , {2y = 5x, {3x + 4y = 5, the solution gives x=5 13, y=25 26 takes the equal sign, so the minimum value is 23 5.

The substitution method is relatively straightforward and easy to think of, you can turn the objective function into a univariate function, and then use the derivative to find the maximum value, but if the objective function is complex and not easy to simplify, it is necessary to solve it with the help of the method of inequality

You're right, it's not true that the inequality is used twice and the conditions are different.

This problem needs to use the basic inequality after the transformation, and the direction of the transformation is to reduce the formula with a reciprocal relation, and then use the basic inequality to reduce the x and y terms to get the minimum value.

Hope it helps!

I think it's 4 and 3 5. I do it by substitution of "1".

Solve 2 firstYou're doing the right thing, (x-2+3)(x-2-3)<0

x+1)(x-5)<0

1You may not understand how to get -1 if(x-5) 0(x+1) 0, then x should be both to the right of 5 on the number line and to the left of -1, then x has no solution.

If (x-5) 0(x+1) 0, then x should be both to the left of 5 and to the right of -1 on the number line, i.e. -1 If (x+1)(x-5) 0, then (x+1) and (x-5) are 0 or 0 at the same time, you can get x 5 or x -1, try it.

You can list a few more inequalities and try it yourself, and you will become proficient, 1Split the inequality into two inequalities: -1 2x squared - x-3 2<-2 and -1 2x squared - x-3 2 -4 (if you mean -1 2 (x squared)).

1 2x squared - x-3 2 -4 is sorted out to obtain x square + 2x-5<0 solution as 1-root number 6 x root number 6-1

1 2x squared - x-3 2 -4 is sorted out to get x square + 2x-1 0 is solved as x root number 2-1 or x 1 - root number 2

The inequality should satisfy both of the above solutions, draw the number axis, and get 1-root number 6 x 1-root number 2 or root number 2-1 x root number 6-1

Sorry, I was careless too, your second question, it was good to use the squared difference formula, but you were wrong.

x squared - 2x - 5<0

x-square-2x+4-9<0

x-2) squared -9<0 The wrong thing here is that x-2x+4 is not equal to (x-2) squared, so don't talk about the next step.

x-2+3)(x-2-3)<0

Sorry so I was wrong too.

x squared - 2x - 5<0

x squared - 2x + 1 - 6<0

x-1) squared -6<0

1.(1) Take x1, x2 [-1,1], and x1 x2, because f(x) is an odd function, so -f(x1)=f(-x1), then f(x2)-f(x1)=f(x2)+f(-x1).

And because f(a)+f(b) a+b 0, then f(x2)+f(-x1) x2-x1 0, because x2-x1 0, so f(x2)+f(-x1) 0, both f(x2)-f(x1) 0 so f(x) is an increasing function on [-1,1].

2) Since f(x) is an increasing function, -1 x+1 2 1 -1 1 x-1 1 x+1 2 1 x-1

lz you can't express it, is it 1 (x-1) or (1 x)-1, so you can figure it out yourself).

2.Vector b point multiplication vector c = (a-1) (x x-2) + (1 (x-2))(2-a) = [x(a-1)+(2-a)] x-2) = [a(x-1)+2-x] (x-2) 1

Both [a(x-1)+2-x] (x-2) -1 0 so [a(x-1)] x-2) 0 and x 1 or x 2 because a is greater than 0

(1) Hypothesis-10

x1-x2<0

So f(x1)1

a-1)(x/x-2)+(1/x-2)(2-a)>1 a-2<0x(a-2)(a-4)/(a-2)

Is the last question premised on the first question?

1.Motonically increasing, take any x1, -x2 on [-1,1], there is f(x1)+f(-x2) x1-x2>0, because f(x) is an odd function, so f(-x2)=-f(x2) then we can know that f(x1)-f(x2) has the same sign as x1-x2, that is, it is an increase function.

Micro increase function, so x+1 2<1 x-1, the rest of the solution inequality landlord should be able to do it himself.

1.Since f(a2-sinx) f(a+3 4+cosx 2) holds true for everything x r.

Again: f(x) is a subtraction function over r.

Then there is: a 2-sinx a+3 4+(cosx) 2, i.e.: a 2-a-3 4 (cosx) 2+sinxa 2-a-3 4 [1-(sinx) 2]+sinx, so:

a^2-a-3/4≥max

Let t=sinx , f(t)=-t 2+t+1 Since x belongs to r, then t=sinx belongs to [-1,1] and f(t)=-(t 2-t-1).

(t^2-t+1/4-5/4)

(t-1/2)^2+5/4

Then when t=1 2, f(t)max=f(1 2)=5 4 Therefore: a 2-a-3 4 max=5 4

Then: a 2-a-2 0

a+1)(a-2)≥0

Then: a 2 or a -1

2.Since for any real number x, there is: x f(x) 1 2(x 2+1).

Let x=1 have: 1 f(1) 1

i.e.: f(1)=1

then from f(-1)=0, we get: a-b+c=0

f(1)=1, obtain: a+b+c=1

Join, get b = a + c = 1 2

And because: for any real number x, there is f(x)-x 0, i.e. ax 2-x 2+c 0

Therefore: the discriminant formula is less than or equal to 0, and a>0

i.e. AC 1 16

And because a+c2 ac2 (1 16)=1 2 and a+c=1 2 is known

So a=c=1 4

Thus there is a real number a=1 4, b=1 2, c=1 4, so that the inequality x f(x) 1 2(x 2+1) holds for the one real number x.

Solution: 2 is not a solution of an inequality, 2a 1, 3 is a solution of an inequality. ∴6a≥1.∴1/6≤a＜1/2.

2 does not belong to A

2a<1

3 belongs to a3a*(3-1) 1

1/6≤a＜1/2

If you have any questions, you can ask me :)