Use Cauchy s inequality to prove a problem master

Proof: 2(a+b+c)[1 (a+b)+1 (b+c)+1 (c+a)].

a+b)+(b+c)+(c+a)][1/(a+b)+1/(b+c)+1/(c+a)]

3[(a+b)(b+c)(c+a)] 1 3) 1 3)=9 divided by (a+b+c) on both sides

Get 2 (a+b) +2 (b+c )+2 (c+a) 9 (a+b+c).

The general proofs of Cauchy's inequality are as follows: The formal way to write Cauchy's inequality is: Remember that the two columns of numbers are ai and bi, then there is.

ai^2)∑bi^2)

∑aibi)^2.

We make f(x).

(ai+xbi)^2(∑bi^2)

x^2+2(∑ai

bi)*x(∑ai^2)

then we know that there is eternity.

f(x)≥0.

With the condition that there is no real root or only one real root of a quadratic function, there is δ=4*

aibi)^2-4

∑ai^2)

∑bi^2)≤0.

So the move came to a conclusion.

Vectors as proof.

m=(a1,a2...an)

n=(b1,b2...bn)

mn=a1b1+a2b2+..anbn=(a1^+a2^+.An ) 1 2 multiplied by (b1 +b2 +.).bn) 1 2 times cosx

Because cosx is less than or equal to 0, so: a1b1+a2b2+.ANBN is less than or equal to A1 + A2 +An ) 1 2 multiplied by (b1 +b2 +.).bn^)^1/2

This proves the inequality. There are many more types of Cauchy inequalities, but here are only two of the more commonly used ones.

Troublesome, according to the formula to allocate the fixed value (cancel the term with x), you should be able to draw conclusions.

Your answer is clearly not logically wrong.

Let's take the simplest example: b+1>=b, b<=1, is b+1 Evergrande more than 1, obviously not necessarily.

Your topic: Analysis: Now that you're using Cauchy, one step is reversed, so we can't use Cauchy's inequality in the first step.

Here's how I did it:

The first step is because yz<=(y 2+z 2) 2,zx<=(x 2+z 2) 2; xy<=(x^2+y^2)/2

In this case, the left side is "=2 3(sigma x 2 (y 2+z 2))).

According to homogeneity, the extraordinary order x 2 + y 2 + z 2 = 1, x 2 = m, y 2 = n, z 2 = r

Then sigma x 2 (y 2 + z 2) = sigma (m (1-m)

According to the convex property of f(x)=x (1-x), the minimum value of sigma(m (1-m) is obtained.

by Cauchy inequality.

1^2+2^2)(x^2+y^2)≥(1*x+2*y)^2=(x+2y)^2=1

x^2+y^2≥1/5

And the landlord should pay attention to x,y 0 in the future.

Substitution x to get 1+4y+5ysquare, when y takes 2 5, the minimum value is 1 5

1 4x square + y square greater than or equal to 2xy

x+2y is greater than or equal to twice the xy in the root number

The two forms can be solved.

Cauchy inequality: For the vector x, there is || = |x||y|If and only if x=y takes equal where.

It is the vector x point multiplied by y (also called inner product, scalar product, quantity product, etc.).

ps: The so-called "product and square < = square and product" in the legend is actually the above one.

Vector x=(a,b,c) y=(b,c,a).

then = ab+bc+ca

x|=(a^2+b^2+c^2)^ y|=(b^2+c^2+a^2)^

So ||= |x||y|

That's it|ab+bc+ca| <= [(a^2+b^2+c^2)^

Because it is a positive number, the absolute value symbol can be removed directly. i.e. a + b + c ab + bc + ca

Take the equivalence as (a,b,c)=(b,c,a) <=> a=b,b=c,c=a <=> a=b=c

PS: The above is the coordinate operation of 3D vectors, which is the generalization of 2D vector coordinate operations in high school.

Addendum: 1A visual illustration of Cauchy's unequal nature:

x||y|cosa a is the included angle.

So apparently there is|| = |x||y|

2.From 1 substituting the coordinate operation, there is "product sum square< = square sum product".

That is, x=(x1,x2) y=(y1,y2) can be generalized to higher dimensions with a two-dimensional explanation.

= |x1y1+x2y2|

x||y|=[(x1)^2+(x2)^2]^ y1)^2+(y2)^2]^

<= |x||y|then |x1y1+x2y2|2 <=(x1) 2+(x2) 2][(y1) 2+(y2) 2] (i.e. "product sum square<= square sum product").

3.Cauchy-Schwartz is a strict proof (a proof in a real product space).

2a+ a^2>= 0

So think of a as an unknown, and the opening of the quadratic function on the right is upward, so delta should be less than or equal to 0

i.e. 4 2-4<= 0

So there is || = |x||y|

4.Proof on the complex inner product space (omitted).

The method is the same as above, and there is mainly a = co-choke.

PS: Hope it helps you in your understanding of Cauchy's inequality.

Open all and can't use Cauchy inequalities directly.

a +b )+1 a) +1 b) ]17 2 first(a +b) (1+1) within (a+b) =1 pushout(a +b) 1 2

Now it is only necessary to prove that (1 a) +1 b) 8 uses the two-tolerant Cauchy inequality (1+1)[(1 a) +1 b) ]1 a+1 b).

With (1 a+1 b)(a+b) (1+1) =4 backwards, you can get (1 a) +1 b) 8 certificates!!

Hopefully, it will enlighten you, be sure to use Cauchy along the conditions for taking the equal sign a=b=1 2.

Do you have to use Cauchy's inequality proof? A +b +c ab+bc+ca is multiplied by 2 on both sides to become (a-b) 2+(b-c) 2+(a-c) 2 0, and the "=" sign is taken if and only if a=b=c.

It is also possible to use the Cauchy inequality, (a + b + c ) (b + c + c ) ab + bc + ca).

Actually, above. Since the method is simpler, I just offer a way to use Cauchy in the hope of enlightening you.