A high school math proof question, a high school math problem, seek proof

One. Consider the function f(x)=x+4 x

f(x1)-f(x2)=(x1x2-4)(x1-x2)/(x1x2)0f(x2)

In the same way, when x1>x2>2, f(x1)>f(x2)two. If you have learned the derivative, it is even simpler, you can know that f(x) decreases monotonically on (0,2) and increases monotonically on (2,+).

a+4 a in 2 to + infinity increments.

Therefore, x1> x2 >2 y1 > y2

A+4 A decreases from 0 to 2.

0< x1 < x2< 2 y1 > y2 is a typical problem.

x+a x(a>0) is incremented from a to infinity at +.

At 0 to the root number a decrement.

f(x)=x+4/x

Finding the derivative of it yields 1-4 x 2

Then when the derivative is greater than zero, f(x) is an increase, and vice versa.

The increase interval is 0 to 2, so when 0< x1 < x2< 2, y1 > y2 decreases from x<0 or x>2, so when x1> x2 >2, y1 > y2

Simple: Subtract 1 from 2, = to the left and right sides of the sign to obtain: y1-y2=4 x1-4 x2, simplified to:

y1-y2=4(x2-x1) x1x2, because 0< x1 < x2< 2, the bottom of the right fraction of the equation is obviously greater than 0, and the upper side is greater than zero, then y1-y2>0, so y1 > y2, the same can be proved when x1> x2 > 2 y1 > y2.

This question actually examines the proof of monotonicity.

Let f(x)=x+4 x, you prove on this basis.

f(x) minus at 0 to 2.

At 2 to positive infinity.

1) n=1.

3+1) 2=2 is divisible.

2) Suppose n=k is divisible by 2. Namely.

3^k+1) /2

n n is a positive integer and 23) dang. n

At k+1 there is (3 n

3^（k+1）

3^k+1) /2

It is also a positive integer.

Proof of 1 n according to (2) (3) (3).

Divisible by 2.

First at n1.

It can't be. 2 to the higher power divisible.

So there is no need to prove it.

Note. Title.

3 to the nth power +1 (n is a positive integer) can be used.

Or. Divisibility of these two propositions holds up by only one proof of one.

In fact, there is only 3 to the odd power of +1

to be divisible by 4.

Of course. There is no need to prove this in this question.

Leave your questions to me!

This question actually examines the proof of monotonicity.

Let f(x)=x+4 x, you prove on this basis.

f(x) minus at 0 to 2.

At 2 to positive infinity.

If you don't understand, ask me again.

This question is wrong, if c=0 is obviously not true.

If C! =0, for a counterexample, a=5 b=1 c=2 d=4

Both sides of the equation are reduced at the same time, c a-b=4 b-d=-3 is not true.

This question is not qualified. c=0 doesn't work.

Is there a range for a, b, c, d?

** Question (is it from the book or left by the teacher)? Which part of the test?

Add a c is not zero, it can't be proven, don't be too complicated Simply give an example of a number and put it in, you will find that there will be a variety of possibilities. It's all high school math and it's still.。。。

The key is to deform the inequality.

Proof is that the original inequality squares the left and right sides:

s=(root(b2(2c-a))) + root(c2(2a-b))) + root(a2(2b-c))))2 (3 abc) 2=9abc

By Cauchy's inequality, (1+1+1)(b 2(2c-a)+c 2(2a-b)+a2(2b-c)))> root number(b2(2c-a))) + root number(c2(2a-b))) + root number(a2(2b-c))))2=s

Therefore, the sufficient conditions for the original inequality to be established are: 9abc>=(1+1+1)(b 2(2c-a)+c 2(2a-b)+a 2(2b-c)).

That is, 3ABC>=B 2(2C-A)+C 2(2A-B)+A2(2B-C)=AB(2A-B)+AC(2C-A)+BC(2B-C).

That is, abc-ab(2a-b) + abc-ac(2c-a) + abc-bc(2b-c) >=0

That is, ab(c+b-2a)+ac(a+b-2c)+bc(a+c-2b)>=0

For any given a,b,c, because the inequality has rotational equivalence of a,b,c.

Consider setting a>=b>=c>=0

then a+b-2c>=0

b+c-2a<=0

Therefore, ac(a+b-2c)>=bc(a+b-2c).

And ab(c+b-2a)+ac(a+b-2c)+bc(a+c-2b)>=ab(c+b-2a)+bc(a+b-2c)+bc(a+c-2b).

ab(c+b-2a)+bc(2a-b-c)=(2a-b-c)(b-a)c>=0

This formula is clearly true.

The original proposition is proven.

The original question is equivalent to proof.

Any x1, x2, x1≠ x2≠1 a

x1-1)/(ax1-1)≠（x2-1)/(ax2-1)(x1-1)/(ax1-1)-(x2-1)/(ax2-1)=[ax1x2-x1-ax2+1-(ax1x2-x2-ax1+1)]/[(ax1-1)(ax2-1)]=[(x2-x1)-a(x2-x1)]/[(ax1-1)(ax2-1)]=(1-a)(x2-x1)/[(ax1-1)(ax2-1)]

A is not equal to 1, x1 is not equal to x2, so the original formula is not equal to 0, so (x1-1) (ax1-1)≠ (x2-1) (ax2-1).

k=(y1-y2) (x1-x2).

k=(1-a)/(ax1-1)(ax2-1)

Since a is not equal to 1, k is not equal to 0, i.e. not parallel to the x-axis.

That is, for any x1≠x2, y1≠y2

y=(x-1) (ax-1) can be converted to y=[(1-a) a 2] (x-1 a)+1 a

This can be obtained by the translation of the inverse proportional function y=[(1-a) a 2] x image, which is still true after translation, since the inverse proportional functions are monotonic in their respective quadrants, and there are always cases where x1 is not equal to x2 and y1 is not equal to y2 for any x1.

The distance from point o to the surface is half of the distance from point A to the surface, so find the distance from point A to the surface first. Find the midpoint E in B1D1, then the distance from A to the surface is the height of the CE side in the triangle ace, according to the geometric relationship, AC= 3, CE=(7) 2 (can be calculated in the triangle CB1D1), AE=CE. In the triangle ace, the height on the ac is 1, and the area of the triangle is, (3) 2, so the height on the CE side is (2 21) 7, then the distance from O to the plane CB1D1 is (21) 7