An inequality proves the conundrum that the individual thinks

Let t=a (a+b),s=b (a+b) then t+s=1y1*y2=(ax1+bx2)(ax2+bx1) (a+b) 2(tx1+sx2)(tx2+sx1)=(t 2+s 2)*x1x2+ts*(x1 2+x2 2).

t^2+s^2)*x1x2+ts*2x1x2(t+s)^2*x1x2=x1x2

So y1y2>=x1x2

The equal sign is obtained when x1 = x2.

Since x1 and x2 are not equal.

So y1y2>=x1x2

y1=(ax1+bx2)/(a+b)

y2=(bx1+ax2)/(a+b)

y1y2=(ax1+bx2)(bx1+ax2)/(a+b)^2[a^2*x1x2+b^2*x1x2+ab(x1^2+x2^2)]/(a+b)^2

a^2+b^2+2ab)*x1x2-2ab*x1x2+ab(x1^2+x2^2)]/(a+b)^2

a+b) 2*x1x2+ab(x1-x2) 2] (a+b) 2x1x2+ab(x1-x2) 2 (a+b) 2 because a, b, x1 , x2 are unequal positive numbers, so x1x2>0, ab>0, so y1y2>x1x2

It's a pleasure to serve you. This question can be used to compare the difference size.

Solution: (a+b) 2*y1y2)-(a+b) 2*x1x2)(ax1+bx2)(bx1+ax2)-(a+b) 2*x1x2)(ab(x1 2+x2 2)+x1x2(a 2+b 2))-x1x2(a 2+b 2)+2abx1x2).

ab(x1^2-2x1x2+x2^2)

ab(x1-x2)^2.

Because a,b,x1,x2 are unequal positive numbers, ab(x1-x2) 2 > 0

So (a+b) 2*y1y2 > a+b) 2*x1x2, so y1y2>x1x2

Since. The denominator is equal. It is used as a difference method.

The denominator of y1-y2=(a-b)(x1-x2) (a+b) is obtained as the positive discussion numerator.

a-b<0 x1-x2<0 or a-b>0 x1-x2>0 is positive, i.e., y1>y2

If there is a difference between two positive and negative, y2 is larger.

If one is equal to zero, then y1=y2

After making a difference score.

It doesn't matter if the denominator is positive.

Multiply the denominator.

ABX1 square + A square x1x2 + B square x1x2 + ABX2 square - (A square x1x2 + B square x1x2 + 2ABX1X2).

Simplify. abx1 square + abx2 square -2abx1x2 proposes ab, because ab is positive, and it doesn't matter, and finally we get a positive number composed of (x1-x2) square multiplied by an ab. Because x1 is not equal to x2, the result is greater than 0

So the former is big.

Let max<, then |1-b|<

So |a+b|,|a-b|At least one of them is greater than b

So max is greater than 1 2

But when a=0, b=, the three formulas are equal to.

It can be proved by counter-proof.

The proof is as follows: if the above three numbers are less than 1 2, there is |1-b|、|a-b|AND |a+b|All are less than 1 2

Then from the first two equations: 1 2b-1 2 can get 1 2<|a+b|<5/2

It contradicts the question, so the question is wrong.

So the original conclusion is correct.

The question is not verified: a + b 1 2

If so, there are many ways to prove it, such as:

Proof 1: Because 1=(a+b) =a +b +2ab a +b +a +b =2(a +b).

So A + B 1 2

Evidence 2: A +B = A +(1-A) =2a -2A +1 = 2(A-1 2) +1 2 1 2

Proof 3: Because a, b belong to (0, positive infinity), and a+b=1, let a=1 2 -x, b=1 2+x

then a +b = (1 2-x) +1 2+x).

1/4-x+x²+1/4+x+x²

1/2+ 2x²≥1/2

Proof-4: A +(1 2) a is obtained from the fundamental inequality

b²+(1/2)²≥b

Add the two formulas to give a +b +1 2 1

i.e. a + b 1 2

Let's discuss whether the proposition can be extended to n terms.

Discussion 1: Expand the number of letters to n.

Let's look at the conditions a > 0,,b > 0 ,c>0 and a+b+c=1, we can prove:a²+b²+c²≥1/3

a²+(1/3)²≥2/3)a

b²+(1/3)²≥2/3)b

c²+(1/3)²≥2/3)c

Add the three formulas to get a + b + c +1 3 2 3

Thus a +b +c 1 3

Promotion: Set A1, A2 ,..., an r+, and a1+a2+....+an=1

then (a1) +a2) +an) 1 n

The proofs are the same as above.

Discussion 2: Expand the number to the nth power.

In the case of conditions a > 0, b > 0 and a+b=1 unchanged,

Proof first: a + b 1 4

a³+(1/2)³+1/2)³≥3•(1/2)•(1/2)•a=(3/4)•a

b³+(1/2)³+1/2)³≥3•(1/2)•(1/2)•b=(3/4)•b

Add the two formulas to get a + b +1 2 3 4

Thus a +b 1 4

Promotion: a n+b n (1 2) (n-1).

This conclusion can be proved by "the arithmetic mean of n positive numbers is not less than their geometric mean".

Upstairs though there were quite a few fights. But it seems that his topic is not like that.

Evidence (a a) * (b b) > = 1 2

The logarithm of both sides is proof: alna+blnb>=ln(1 2).

Constructor f(x)=xlnx

f''(x)=1 x>0 so f(x) is convex, which is obtained by the property of the convex function (i.e., the inequality of piano birth).

f(a)+f(b)]/2>=f[(a+b)/2]

i.e. (alna+blnb) 2>(1 2)*ln(1 2) is thus proven.

The same idea is applied to N-yuan, AI >0, i=1, 2, 3....and a1+a2+.an=1

Certificate: (A1 A1) (A2 A2)...an^an)>=1/n

Take the logarithm to prove a1lna+a2lna2+.anlnan>=ln(1/n)

Construct f(x)=xlnx because f(x) is convex.

Thus [f(a1)+.].f(an)]/n>=f[(a1+a2+..an)/n]=(1/n)*ln(1/n)

So f(a1)+f(a2)+f(an)>=ln(1/n)

Therefore, it has been proved that it is exactly the same as the idea of the 2-yuan form.

It is not an image of traditional performances, and it will also cause some students to follow suit, which will backfire.

I suggest that you use dance as a form of performance to visualize it, with less violence and more artistry. I recommend you that the background ** for the MJ of this song is to reflect this aspect of the problem, you can refer to the MV of the song, start with a fierce dance, divide the actors into two groups, one group is weak, one group is strong, they sing in duet, the body can have some contact, to express the situation from the eyes, so the performance and capture of the eyes is very important, after the intense dance, you should change to a more melancholy **, the dance should be soothing, and the eyes should be full of thinking, to express repentance and reflection on violence. The final ** background should be cheerful, the two groups of people should dance friendly, mainly emphasize body language, and help in harmony.

There is a mistake in this question.

For example, take a=11 10, b=1, c=19 21; Then ab+bc+ca=3, but.

a 2 + b 2 + c 3 + 3 abc = approximately equal to not satisfying 6;So the title must be wrong.

The title should look like this:

Knowing that a, b, and c are positive real numbers, and ab+bc+ca=3, it is correct to verify a 3+b 3+c 3+3abc 6".

The proof of this question will be added later.

Are you sure you wrote the verification form correctly?

(1) ......... by the basic inequality: a+b>=2 ab1b+c>=2√bc………2

a+c>=2√ac………3

So. Add 1 2 3 to get AB+ BC+ AC<=1 2 and prove it again》=1 3: (I didn't expect it for the time being).

2) (2) Use the Cauchy inequality proof ( a 2 + b 2 + c 2) (1 + 1 + 1) > = (a + b + c) 2, and that's it.

The proposition is wrong:

A 0, B 0, C 1, Question 1 0, False proposition.

by mean inequalities.

x²+1>=2x

y²+2>=2√(2*y²)=2√y

z²+8>=2√8z²=4√2z

Multiply (x +1) (y + 2) (z + 8) > = 32xyz

a2 (a-b) + (a-b) > 2a (basic inequalities) b2 (b-c) + (b-c) >2b

So a2 (a-b)+(a-b)+b2(b-c)+(b-c)>2a+2b

So a2 (a-b)+b2 (b-c)>a+2b+c

Let x=sina, y=cosa, x>=0, i.e., a [0, ]x+y=sina+cosa= 2sin(a+ 4) because 4<=a+ 4<=5 4

Therefore, the original formula has a minimum value of -1 at a=

At this point, x=0 and y=-1

x=√(1-y^2)

So y 2 1

Y [-1,1] is obtained

Therefore x belongs to [0,1].

Let z=x+y, then the original problem is to find the range of z.

From the idea of combining numbers and shapes, the problem can be regarded as the intersection of a straight line (z=x+y) and a half-circle (x= (1-y 2)).

So when y=-1, x=0 is obtained from x= (1-y 2), at this time zmin=-1, when y= 2 2, x= 2 2, at this time zmax= 2 In summary, the minimum value of x+y is -1

Let y=sina then x=under the root sign (1-y 2) x=cosa (0《a《 )x+y=sina+cosa The minimum value is a= when y=-1 x=0 -1 thank you.

It can be used analytically:

ax+by)/(a+b)>(x+y)/2<=2ax+2by>(x+y)(a+b)

2ax+2by>ax+bx+ay+by<=ax+by>ay+bx

ax-ay>bx-by

a(x-y)>b(x-y)

Because x>y, x-y>0

a>b

This is clearly true.

Suppose (ax+by) (a+b)=(x+y) 22(ax+by)=(a+b)(x+y)=ax+by+bx+aybx+ay=0

Since a, b, x, y, are all greater than 0, bx+ay>0, so the assumption is not true.

Proof : To prove (ax + by) (a+b) x+y) 2 only need to prove (ax +by) (a+b) -x+y) 2 0 because a, b, x, y are all greater than zero.

Multiply both sides of the inequality by 2(a+b), and the invariant sign of the inequality gives 2(ax+by)- a+b)(x+y) 0 to (a-b)(x-y) 0

Set a b, x y

Obviously a-b 0, x-y 0

Hence the inequality holds. Certification.