Several Proofs of Cauchy s Inequality detailed .

Method 1: Vector analysis.

Let a=(a1,a2,......an）b=（b1，b2，……bn）

a·b≤|a||b|

a1b1+a2b2+……anbn≤√a1²+a2²+…an²√b1²+b2²+…bn²

aibi）²≤ai²σbi²

Method 2: Construct a quadratic function.

When a1=a2=....=an=0 or b1=b2=....=bn=0, the general form is clearly true.

Let a = ai 2 b = ai · bi c = bi 2

When a1=a2=...., when at least one of an is not zero, we know that a>0

Constructing the quadratic function f(x)=ax 2+2bx+c, gets:

f(x)=∑(ai^2·x^2+2ai·bi·x+bi^2)=∑ (ai·x+bi)^2≥0

Therefore, the discriminant formula of f(x) = 4b 2 4ac 0, and the shift gives ac b 2, and the inequality has been proved.

Proof in general form.

Proof: ( ai 2)( bi 2) ai·bi) 2 Prove: When a1=a2=...=an=0 or b1=b2=....=bn=0, the general form is clearly true so that a= ai 2 b= ai·bi c= bi 2 when a1, a2 ,..., when at least one of the AN is not zero, we know that a>0 constructs the quadratic function f(x)=ax 2+2bx+c, (note that the coefficient of the primary term is 2b, not b) to obtain:

f(x)= (ai 2·x 2+2ai·bi·x+bi 2)= (ai·x+bi) 2 0 Therefore, the discriminant formula of f(x) =4b 2 4ac 0, (Please note: the discriminant formula of the unary quadratic equation ax 2+bx+c=0 is indeed =b 2-4ac, but the equation ax 2+2bx+c = 0 has been replaced as follows: a = a, b = 2b, c = c, where b has been replaced with 2b, which has led to the misunderstanding of many netizens.) If this step is wrong, the Cauchy inequality cannot be proven!

Shift the term to obtain ACB 2, and the inequality has been proved.

Proof in vector form.

Let m=(a1, a2, ...an)，n=(b1, b2, …bn) m·n=a1b1+a2b2+…+anbn=|m||n|cos=√(a1^2+a2^2+…+an^2) ×b1^2+b2^2+…+bn^2) ×cos∵cos≤1 ∴a1b1+a2b2+…+anbn≤√(a1^2+a2^2+…+an^2) ×b1^2+b2^2+…+bn 2) Note: "" denotes the square root.

I think of quadratic function construction, Lagrange identities and mathematical induction, the process is too long, go and check the relevant information yourself.

Cauchy inequality: ai, bi r, verify: (a1 2+a2 2+..

The Cauchy inequality was obtained by the great mathematician Cauchy when he was working on the problem of "flow numbers" in mathematical analysis. Historically, however, this inequality should be called the Cauchy-Buniakowsky-Schwarz inequality, because it was the latter two mathematicians, independently of each other, who generalized it in integralism, that led to the near-perfection of the inequality.

Cauchy inequality is an inequality discovered by Cauchy in the research process, which has a very wide application in solving the relevant problems of inequality proof, so it is very important in the improvement and research of higher mathematics, and is one of the research contents of higher mathematics.

Cauchy Augustin-Louis (1789-1857), a French mathematician, was born on August 21, 1789 in Paris, the son of his father, Louis François Cauchy, a French Bourbon who held public office in the turbulent political maelstrom of France. For family reasons, Cauchy himself belonged to the orthodox faction that embraced the Bourbons and was a devout Catholic.

The proof of Cauchy's inequality is:

If the number of the two columns is ai, bi, then there is (ai 2)*(bi 2) (ai*bi) 2.

Let f(x)= ai+x*bi) 2=( bi 2)*x 2+2*( ai*bi)*x+( ai 2), then there is always f(x) 0.

With the condition that the quadratic function has no real root or only one real root, there is δ=4*( ai*bi) 2-4*( ai 2)*(bi 2) 0, so the conclusion is obtained by shifting.

Cauchy inequality is an inequality discovered by Cauchy in the research process, which has a very wide application in solving the relevant problems of inequality proof, so it is very important in the improvement and research of higher mathematics, and is one of the research contents of higher mathematics.

It is said that when the Journal of the French Academy of Sciences was founded, there were so many works by Cauchy that the Academy of Sciences had to bear a lot of printing costs, which exceeded the budget of the Academy of Sciences, so the Academy of Sciences later stipulated that the longest could only be four pages. Cauchy's longer ** had to be submitted elsewhere.

1. Two-dimensional form.

Formula deformation: <>

2. Vector form.

3. Triangular form.

4. Probability theory forms.

5. Integral form.

Cauchy inequality: ai, bi r, verify: (a1 2+a2 2+..an^2)*(b1^2+b2^2+..bn^2)≥(a1*b1+a2*b2+..an*bn)^2.

I think the easiest way is to construct n-dimensional vectors: =a1,a2,..an)，βb1,b2,..bn).

then (a1 2+a2 2+..an^2)*√b1^2+b2^2+..bn^2)=|cos<α,a1*b1+a2*b2+..an*bn.

Both sides are squared at the same time: (a1 2+a2 2+..an^2)*(b1^2+b2^2+..bn^2)≥(a1*b1+a2*b2+..an*bn)^2.

There are many other methods: the combination of numbers and shapes:

The formal way to write Cauchy's inequality is: remember that the two columns of numbers are ai, bi, and there is.

ai^2) *bi^2) ≥ai * bi)^2.

We order. f(x) =ai + x * bi)^2

bi^2) *x^2 + 2 * ai * bi) *x + ai^2)

then we know that there is eternity.

f(x) ≥0.

With quadratic functions, there is no real root or only one real root condition.

4 * ai * bi)^2 - 4 * ai^2) *bi^2) ≤0.

So the move came to a conclusion.

In addition to this, there is also a way to make errands. Wait a minute.

As mentioned above, it can be proved as an auxiliary quadratic function.

Actually, it's simpler, use vectors to prove it.

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