Khác biệt giữa bản sửa đổi của “Bất đẳng thức trung bình cộng và trung bình nhân”
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Dòng 40:
Đẳng thức khi và chỉ khi <math>x_1 = x_2 = \cdots = x_n</math>
==Ví dụ dụng dụng==
Cho hàm số sau:
:<math>f(x,y,z) = \frac{x}{y} + \sqrt{\frac{y}{z}} + \sqrt[3]{\frac{z}{x}}</math>
Với ''x'', ''y'' và ''z'' là các số thực dương. Giả sử rằng ta phải tìm giá trị nhỏ nhất của hàm số đã cho. Biến đổi và áp dụng ''bất đẳng thức Cauchy'' ta có:
:{|
|<math>f(x,y,z)\,\;</math> || <math>= 6 \cdot \frac{ \frac{x}{y} + \frac{1}{2} \sqrt{\frac{y}{z}} + \frac{1}{2} \sqrt{\frac{y}{z}} + \frac{1}{3} \sqrt[3]{\frac{z}{x}} + \frac{1}{3} \sqrt[3]{\frac{z}{x}} + \frac{1}{3} \sqrt[3]{\frac{z}{x}} }{6}</math>
|-
| || <!-- add a bit of space -->
|-
| || <math>\ge 6 \cdot \sqrt[6]{ \frac{x}{y} \cdot \frac{1}{2} \sqrt{\frac{y}{z}} \cdot \frac{1}{2} \sqrt{\frac{y}{z}} \cdot \frac{1}{3} \sqrt[3]{\frac{z}{x}} \cdot \frac{1}{3} \sqrt[3]{\frac{z}{x}} \cdot \frac{1}{3} \sqrt[3]{\frac{z}{x}} }</math>
|-
| || <!-- add a bit of space -->
|-
| || <math>= 6 \cdot \sqrt[6]{ \frac{1}{2 \cdot 2 \cdot 3 \cdot 3 \cdot 3} \frac{x}{y} \frac{y}{z} \frac{z}{x} }</math>
|-
| || <!-- add a bit of space -->
|-
| || <math>= 2^{2/3} \cdot 3^{1/2}</math>
|}
Vậy ta có giá trị nhỏ nhất của:
:<math>f(x,y,z) \mbox{là} 2^{2/3} \cdot 3^{1/2} \quad \mbox{khi} \quad \frac{x}{y} = \frac{1}{2} \sqrt{\frac{y}{z}} = \frac{1}{3} \sqrt[3]{\frac{z}{x}}.</math>
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==Chứng minh bằng quy nạp==
Đặt:
:<math>\mu=\frac{\ x_1 + \cdots + x_n}n</math>
bất đăng thức tương đương với<br />
''x''<sub>1</sub>,...,''x<sub>n</sub>'' là các [[số thực]] không âm, ta có:
:<math>\mu^n\ge x_1 x_2 \cdots x_n\,</math>
dấu bằng xãy ra nếu ''μ'' = ''x<sub>i</sub>'' với mọi ''i'' = 1,...,''n''.
Chứng minh dưới đây áp dụng phương pháp [[quy nạp toán học].
'''Cơ sở''' Với ''n'' = 1 bất đẳng thức đúng.
'''Giả thiết quy nạp:''' Giả sử rằng bất đẳng thức đúng với n (n lớn hơn hoặc bằng 1).
'''Quy nạp:''' Xét''n'' + 1 số thực không âm. Ta có
:<math> (n+1)\mu=\ x_1 + \cdots + x_n + x_{n+1}.\,</math>
Nếu tất cả các số đều bằng ''μ'', thì ta có đẳng thức và coi như xong. Ngược lại ta sẽ tìm được ít nhất một số nhỏ hơn ''μ'' và một số lớn hơn ''μ'', không mất tính tổng quát, xem rằng ''x<sub>n</sub>'' > ''μ'' và ''x''<sub>''n''+1</sub> < ''μ''. Then
:<math>(x_n-\mu)(\mu-x_{n+1})>0\,.\qquad(*)</math>
Now consider the ''n'' numbers
:<math>x_1, \ldots, x_{n-1}, x_n'</math> with <math>x_n':=x_n+x_{n+1}-\mu\ge x_n-\mu>0\,,</math>
which are also non-negative. Since
:<math>n\mu=x_1 + \cdots + x_{n-1} + \underbrace{x_n+x_{n+1}-\mu}_{=\,x_n'},</math>
''μ'' is also the arithmetic mean of <math>x_1, \ldots, x_{n-1}, x_n'</math> and the induction hypothesis implies
:<math>\mu^{n+1}=\mu^n\cdot\mu\ge x_1x_2 \cdots x_{n-1} x_n'\mu.\qquad(**)</math>
Due to (*) we know that
:<math>(\underbrace{x_n+x_{n+1}-\mu}_{=\,x_n'})\mu-x_nx_{n+1}=(x_n-\mu)(\mu-x_{n+1})>0,</math>
hence
:<math>x_n'\mu>x_nx_{n+1}\,,\qquad({*}{*}{*})</math>
in particular ''μ'' > 0. Therefore, if at least one of the numbers ''x''<sub>1</sub>,...,''x''<sub>''n''−1</sub> is zero, then we already have strict inequality in (**). Otherwise the right-hand side of (**) is positive and strict inequality is obtained by using the estimate (***). Therefore, substituting (***) into (**) gives in both cases
:<math>\mu^{n+1}>x_1x_2 \cdots x_{n-1} x_nx_{n+1}\,,</math>
which completes the proof.
==Chứng mình của Pólya==
[[George Pólya]] provided the following proof, using the [[exponential function]] and the inequality ''e''<sup>''x''</sup> ≥ 1 + ''x'', which is valid for every real number ''x''. To verify this inequality, note that both sides as well as their first [[derivative]]s agree for ''x'' = 0 and that the exponential function is [[convex function|strictly convex]], because its second derivative is positive for every real ''x''. For this reason, also the equality ''e''<sup>''x''</sup> = 1 + ''x'' only holds for ''x'' = 0.
Let μ be the arithmetic mean, and let ρ be the geometric mean of ''x''<sub>1</sub>, ..., ''x''<sub>''n''</sub>. If all ''x''<sub>1</sub>, ..., ''x''<sub>''n''</sub> are equal, then μ = ρ.
It remains to prove the strict inequality μ > ρ if ''x''<sub>1</sub>, ..., ''x''<sub>''n''</sub> ≥ 0 are not all equal. Then, in particular, they are not all zero, hence μ > 0.
If we substitute ''x''<sub>''i''</sub> /μ − 1 for ''x'' in the above inequality ''e''<sup>''x''</sup> ≥ 1 + ''x'' we get that
:<math>\exp\Bigl({x_i \over \mu} - 1\Bigr)\ge {x_i \over \mu}\,</math>
for each ''i'' and strict inequality for those ''i'' with ''x''<sub>''i''</sub> ≠ μ. Since ''x''<sub>''i''</sub> /μ ≥ 0, we can multiply all these inequalities together, side by side, for ''i'' = 1, ..., ''n'', to obtain
:<math>\prod_{i=1}^n\exp\Bigl({x_i \over \mu} - 1\Bigr) > \prod_{i=1}^n {x_i \over \mu}\,,</math>
where we get strict inequality because no factor on the left-hand side is zero and there was strict inequality for at least one ''i''. Using the [[functional equation]] of the exponential function, we get
:<math>\exp\biggl(\frac1\mu\sum_{i=1}^n x_i - n\biggr) > \prod_{i=1}^n {x_i \over \mu}\,.\qquad(*)</math>
Since μ is the arithmetic mean, the summation in the parentheses on the left of (*) can be reduced to
:<math>\sum_{i=1}^n x_i = n\mu\,.</math>
Thus, the left-hand side of the inequality (*) is exp(''n'' − ''n'') = 1. Since ρ is the geometric mean, the product on the right of (*) can be rewritten as
:<math>\frac1{\mu^n}\prod_{i=1}^n x_i = {\rho^n \over \mu^n}.</math>
So (*) reduces to 1 > ρ<sup>''n''</sup>/μ<sup>''n''</sup> and hence μ > ρ.
== Chứng minh của Cauchy ==
The following proof by cases relies directly on well-known rules of arithmetic but employs the rarely-used technique of forward-backward-induction. It is essentially from [[Augustin Louis Cauchy]] and can be found in his ''Cours d'analyse''.
=== The case where all the terms are equal ===
If all the terms are equal:
:<math>x_1 = x_2 = \cdots = x_n</math>
then their sum is ''nx''<sub>1</sub>, so their arithmetic mean is ''x''<sub>1</sub>; and their product is ''x''<sub>1</sub><sup>''n''</sup>, so their geometric mean is ''x''<sub>1</sub>; therefore, the arithmetic mean and geometric mean are equal, as desired.
=== The case where not all the terms are equal ===
It remains to show that if ''not'' all the terms are equal, then the arithmetic mean is greater than the geometric mean. Clearly, this is only possible when ''n'' > 1.
This case is significantly more complex, and we divide it into subcases.
==== The subcase where ''n'' <nowiki>=</nowiki> 2 ====
If ''n'' = 2, then we have two terms, ''x''<sub>1</sub> and ''x''<sub>2</sub>, and since (by our assumption) not all terms are equal, we have:
: <math>
\begin{align}
x_1 & \ne x_2 \\[3pt]
x_1 - x_2 & \ne 0 \\[3pt]
\left( x_1 - x_2 \right) ^2 & > 0 \\[3pt]
x_1^2 - 2 x_1 x_2 + x_2^2 & > 0 \\[3pt]
x_1^2 + 2 x_1 x_2 + x_2^2 & > 4 x_1 x_2 \\[3pt]
\left( x_1 + x_2 \right) ^2& > 4 x_1 x_2 \\[3pt]
\Bigl( \frac{x_1 + x_2}{2} \Bigr)^2 & > x_1 x_2 \\[3pt]
\frac{x_1 + x_2}{2} & > \sqrt{x_1 x_2}
\end{align}
</math>
as desired.
==== The subcase where ''n'' <nowiki>=</nowiki> 2<sup>''k''</sup> ====
Consider the case where ''n'' = 2<sup>''k''</sup>, where ''k'' is a positive integer. We proceed by [[mathematical induction]].
In the base case, ''k'' = 1, so ''n'' = 2. We have already shown that the inequality holds where ''n'' = 2, so we are done.
Now, suppose that for a given ''k'' > 1, we have already shown that the inequality holds for <span style="white-space:nowrap">''n'' = 2<sup>''k''−1</sup></span>, and we wish to show that it holds for <span style="white-space:nowrap">''n'' = 2<sup>''k''</sup></span>. To do so, we proceed as follows:
: <math>
\begin{align}
\frac{x_1 + x_2 + \cdots + x_{2^k}}{2^k} & {} =\frac{\frac{x_1 + x_2 + \cdots + x_{2^{k-1}}}{2^{k-1}} + \frac{x_{2^{k-1} + 1} + x_{2^{k-1} + 2} + \cdots + x_{2^k}}{2^{k-1}}}{2} \\[7pt]
& \ge \frac{\sqrt[2^{k-1}]{x_1 \cdot x_2 \cdots x_{2^{k-1}}} + \sqrt[2^{k-1}]{x_{2^{k-1} + 1} \cdot x_{2^{k-1} + 2} \cdots x_{2^k}}}{2} \\[7pt]
& \ge \sqrt{\sqrt[2^{k-1}]{x_1 \cdot x_2 \cdots x_{2^{k-1}}} \cdot \sqrt[2^{k-1}]{x_{2^{k-1} + 1} \cdot x_{2^{k-1} + 2} \cdots x_{2^k}}} \\[7pt]
& = \sqrt[2^k]{x_1 \cdot x_2 \cdots x_{2^k}}
\end{align}
</math>
where in the first inequality, the two sides are equal only if both of the following are true:
:<math>x_1 = x_2 = \cdots = x_{2^{k-1}}</math>
:<math>x_{2^{k-1}+1} = x_{2^{k-1}+2} = \cdots = x_{2^k}</math>
(in which case the first arithmetic mean and first geometric mean are both equal to ''x''<sub>1</sub>, and similarly with the second arithmetic mean and second geometric mean); and in the second inequality, the two sides are only equal if the two geometric means are equal. Since not all 2<sup>''k''</sup> numbers are equal, it is not possible for both inequalities to be equalities, so we know that:
:<math>\frac{x_1 + x_2 + \cdots + x_{2^k}}{2^k} > \sqrt[2^k]{x_1 \cdot x_2 \cdots x_{2^k}}</math>
as desired.
==== The subcase where ''n'' < 2<sup>''k''</sup> ====
If ''n'' is not a natural power of 2, then it is certainly ''less'' than some natural power of 2, since the sequence <span style="white-space:nowrap">2, 4, 8, . . ., 2<sup>''k''</sup>, . . .</span> is unbounded above. Therefore, without loss of generality, let ''m'' be some natural power of 2 that is greater than ''n''.
So, if we have ''n'' terms, then let us denote their arithmetic mean by α, and expand our list of terms thus:
:<math>x_{n+1} = x_{n+2} = \cdots = x_m = \alpha.</math>
We then have:
: <math>
\begin{align}
\alpha & = \frac{x_1 + x_2 + \cdots + x_n}{n} \\[6pt]
& = \frac{\frac{m}{n} \left( x_1 + x_2 + \cdots + x_n \right)}{m} \\[6pt]
& = \frac{x_1 + x_2 + \cdots + x_n + \frac{m-n}{n} \left( x_1 + x_2 + \cdots + x_n \right)}{m} \\[6pt]
& = \frac{x_1 + x_2 + \cdots + x_n + \left( m-n \right) \alpha}{m} \\[6pt]
& = \frac{x_1 + x_2 + \cdots + x_n + x_{n+1} + \cdots + x_m}{m} \\[6pt]
& > \sqrt[m]{x_1 \cdot x_2 \cdots x_n \cdot x_{n+1} \cdots x_m} \\[6pt]
& = \sqrt[m]{x_1 \cdot x_2 \cdots x_n \cdot \alpha^{m-n}}\,,
\end{align}
</math>
so
: <math>
\begin{align}
\alpha^m & > x_1 \cdot x_2 \cdots x_n \cdot \alpha^{m-n} \\[3pt]
\alpha^n & > x_1 \cdot x_2 \cdots x_n \\[3pt]
\alpha & > \sqrt[n]{x_1 \cdot x_2 \cdots x_n}
\end{align}
</math>
as desired.
==Proof of the generalized AM-GM inequality using Jensen's inequality==
Using the finite form of [[Jensen's inequality]] for the [[natural logarithm]], we can prove the inequality between the weighted arithmetic mean and the weighted geometric mean stated above.
Since an ''x<sub>k</sub>'' with weight ''α<sub>k</sub>'' = 0 has no influence on the inequality, we may assume in the following that all weights are positive. If all ''x<sub>k</sub>'' are equal, then equality holds. Therefore, it remains to prove strict inequality if they are not all equal, which we will assume in the following, too. If at least one ''x<sub>k</sub>'' is zero (but not all), then the weighted geometric mean is zero, while the weighted arithmetic mean is positive, hence strict inequality holds. Therefore, we may assume also that all ''x<sub>k</sub>'' are positive.
Since the natural logarithm is [[concave function|strictly concave]], the finite form of Jensen's inequality and the [[functional equation]]s of the natural logarithm imply
:<math>
\ln\biggl(\frac{\alpha_1x_1+\cdots+\alpha_nx_n}\alpha\biggr)
>\frac{\alpha_1}\alpha\ln x_1+\cdots+\frac{\alpha_n}\alpha\ln x_n
=\ln \sqrt[\alpha]{x_1^{\alpha_1} x_2^{\alpha_2} \cdots x_n^{\alpha_n}}.
</math>
Since the natural logarithm is [[monotonic function|strictly increasing]],
:<math>
\frac{\alpha_1x_1+\cdots+\alpha_nx_n}\alpha
>\sqrt[\alpha]{x_1^{\alpha_1} x_2^{\alpha_2} \cdots x_n^{\alpha_n}}.
</math>
==Ứng dụng trong lí thuyết toán==
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