Mathematics for the second year of high school Proof and application of inequalities super difficul

Okay, okay, I'll do it if I look at the landlord so pitiful.,Actually, these questions are not difficult.,The proof process is no more than 5 lines.,Click on the idea first.,If you don't look at the answer in my ** again.。

1.Cauchy inequality or sorting inequality.

2.It is easiest to use the Cauchy inequality and then use the basic inequality ab+bc+ca<=a 2+b 2+c 2.

3.Scaling with the mean is sufficient.

4.First the element, then the Cauchy inequality, and finally the mean.

5.Just use the mean.

In addition, my extreme bs: 2009-8-4 12:25 arrogant behavior, you are so bullish will not answer yourself! Why are you still hiding and not standing up? I'm the most bss of this kind of levelless and narrow-minded person!

You're really talented, that's Cauchy inequality, and sorting inequality you wait a minute, I'll give you the order out of order and reverse order right away.

x1x2/x1+x2x3/x2+..x1xn/x1x1+x2+x3+..xn

a^4/bc+b^4/ac+c^4/ab

a^2+b^c+c^2

You garbage, the first question you use the sorting inequality to do it, wrong! The first problem is rotational symmetry, not perfect symmetry, you can't assume size, the correct method is Cauchy inequality! If you must use a sort inequality, you need to transform

1.Direct Formulation.

The original inequality is equivalent to (a-2) 2+(b+1) 2 0, and this is clearly true.

2.Multiply 2 recipes on both sides.

Equivalent to (a-b) 2+(b-c) 2+(c-a) 2 0 This is clearly true.

1.It is used to compare differences, move all the terms on the right to the left, find a way to prove that the sum of the left sides of the inequality is greater than or equal to zero, and match them into two perfect square terms of (a-2) 2 and (b+1) 2.

2.It's also a bad comparison, just match them into 3 perfectly square terms, here's a little trick, multiply the left and right sides of the inequality by 2.

The purpose of this square class is to match the perfect square term as much as possible, and to prove it by its property that it is equal to 0.

1.(a-2) +b+1) =a -4a+4+b +2b+1 0 move the terms, merge the same kind of terms.

i.e. a + b + 5 2 (2a-b).

2.(a-b) +b-c) +c-a) =2a +2b +2c -2ab-2bc-2ca 0 shift term, remove 2

i.e. a + b + c ab + bc + ac

Use the mean inequality (the arithmetic mean is greater than the geometric mean) and multiply the left and right parts to get the right. Try it yourself, it's simple. (The mean inequality of the four terms: a+b+c+d is greater than or equal to 4 fourth-degree root number ABCD).

Proof of: (ab+a+b+1) (ab+ac+bc+c) 4 square a square b under the fourth root number 4 square a square b square c square 4 square b square c square under the fourth root number 16 FOURTH POWER A 4th power b 4th power C 4th power = 16abc.

The left side is minus the right side is greater than zero.

In the title, n is sufficiently large, you can take n=([a] 1) Note: [a] represents the whole part of a (a 0 is assumed here). Then 0 a 2 ([a] 1) 2. And so it proves out.

The scope of such a reduction is difficult to grasp, and there is often a phenomenon that no conclusion can be reached after the reduction, so it should be replaced by other methods.

1: Is there a constant c such that the inequality (x 2x+y) + (y x+2y) "c (x x+2y)+(y 2x+y), is constant for any positive real number x,y? Prove your conclusions.

2: Verify 1 (n+1)*[1+1 3+1 5+......1/(2n-1)]>1/n*(1/2+1/4+1/6+……1/2n),n>2

3: a, b, c are all real numbers.

If a+b+c=1

Verification: A square + B square + C square is greater than or equal to 1 3

That information might be a little helpful, if you want to know more, you might as well take a look at these two:

Proof of inequality [There are e-book methods and techniques for solving inequalities.

…I can't remember the few inequalities mentioned, I can't tell them apart, can you add them?

Using the Lagrangian theorem, it's simple.

Knowing the positive real number a ≠ b, verify: a b+b a > a+ b

Proof: a b+b a- a- b = a-b) b-(a-b) a = a-b)( a- group b) ab).

If a > b then a-b > 0, a- b > 0, get (a-b) ( a- b) ab) >0

If a < b then the old a-b < 0, a-b < 0, the same (a-b)( a- b) ab) >0

So a b+b a- a- b > 0, i.e. a b+b a > a+ b

Another method is to follow the sliding to continue to deform to obtain: (a-b)( a- b) ab) =a-b) a b+b a) >0

1. It can be seen that a and b must be greater than 0, both sides are squared at the same time, eliminate 2 times the root number ab2, and multiply both sides by the spike ab at the same time to obtain a cubic plus b cubic greater than a square b plus b square a3, and shift the common factor to obtain a square * (a-b) > b square * (b-a) 4, assuming a > b, the guess is true.

Assuming b>a, the formula is also true.

Let f(x) = x 3 - 2x 2 + 2x - 1 then f(x) = x 3 - 2x 2 + 2x - 1 = (x 2-x+1)(x-1).

Because x 2-x+1=(x-1 2) 2+3 4>0 and from known x 1 we get x-1 0

So (x 2-x+1)(x-1) 0

i.e. x 2x -2x+1

Just prove x -x +x x -x+1, i.e. x(x -x+1) x -x+1;

It is known that x 1 is shifted to get x 2x -2x+1.

x³-2x²+2x-1

x³-2x²+x+x-1

x(x²-2x+1)+(x-1)

x(x-1)^2+(x-1)

x-1)[(x(x-1)+1]

The above terms are all non-negative numbers, so the above formula is non-negative.

That is, the original formula is proven.