Difference between revisions of "SFMT"
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Given nonnegative real numbers <math>e_1, e_2, e_3, \cdots ,e_n</math> and that <math>x_1+x_2+x_3+\cdots+x_n</math> is fixed and all the terms inside the sum are nonnegative, the maximum value of <math>x_1^{e_1}x_2^{e_2}x_3^{e_3}\cdots x_n^{e_n}</math> is when <math>\dfrac{x_1}{e_1}=\dfrac{x_2}{e_2}=\dfrac{x_3}{e_3}=\cdots =\dfrac{x_n}{e_n}.</math> | Given nonnegative real numbers <math>e_1, e_2, e_3, \cdots ,e_n</math> and that <math>x_1+x_2+x_3+\cdots+x_n</math> is fixed and all the terms inside the sum are nonnegative, the maximum value of <math>x_1^{e_1}x_2^{e_2}x_3^{e_3}\cdots x_n^{e_n}</math> is when <math>\dfrac{x_1}{e_1}=\dfrac{x_2}{e_2}=\dfrac{x_3}{e_3}=\cdots =\dfrac{x_n}{e_n}.</math> | ||
== Proof == | == Proof == | ||
− | The weighted | + | The weighted AM-GM Inequality states that if <math>a_1, a_2, \dotsc, a_n</math> are nonnegative real numbers, and <math>\lambda_1, \lambda_2, \dotsc, \lambda_n</math> are nonnegative real numbers (the "weights") which sum to 1, then <cmath>\lambda_1 a_1 + \lambda_2 a_2 + \dotsb + \lambda_n a_n \ge a_1^{\lambda_1} a_2^{\lambda_2} \dotsm a_n^{\lambda_n}.</cmath> We let <math>\lambda_i=\dfrac{e_i}{e_1+e_2+\cdots+e_n}</math> and <math>a_i=\dfrac{x_i}{e_i}</math> in this inequality. We get that <math>x_1+x_2+x_3+\cdots+x_n \ge \sqrt[n]{\dfrac{x_1^{e_1}x_2^{e_2}x_3^{e_3}\cdots x_n^{e_n}}{e_1^{e_1}e_2^{e_2}e_3^{e_3}\cdots e_n^{e_n}}}(e_1+e_2+e_3+\cdots+e_n).</math> Dividing both sides and then taking to the nth power, we get <math>\left(\dfrac{x_1+x_2+x_3+\cdots+x_n}{e_1+e_2+e_3+\cdots+e_n}\right)^n \ge \dfrac{x_1^{e_1}x_2^{e_2}x_3^{e_3}\cdots x_n^{e_n}}{e_1^{e_1}e_2^{e_2}e_3^{e_3}\cdots e_n^{e_n}}.</math>Then we can multiply both sides to get <math>x_1^{e_1}x_2^{e_2}x_3^{e_3}\cdots x_n^{e_n} \le \left(\dfrac{x_1+x_2+x_3+\cdots+x_n}{e_1+e_2+e_3+\cdots+e_n}\right)^n (e_1^{e_1}e_2^{e_2}e_3^{e_3}\cdots e_n^{e_n}).</math> The equality case for the weighted AM-GM inequality is when all the <math>a_i</math> terms such that <math>e_i</math> is not <math>0</math> are equal, or in this case <math>\dfrac{x_1}{e_1}=\dfrac{x_2}{e_2}=\dfrac{x_3}{e_3}=\cdots=\dfrac{x_n}{e_n}.</math> |
+ | |||
+ | == Example Problems == | ||
+ | === Problem 1 === | ||
+ | Given nonnegative integers <math>x, y, z</math> such that <math>x+y+z=16,</math> find the maximum value <math>x^3y^2z^3.</math> | ||
+ | |||
+ | === Solution 1 === | ||
+ | By the new theorem, we know that <math>\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{3},</math> so <math>x=6, y=4, z=6.</math> Plugging it in, our answer is <math>746496.</math> | ||
+ | |||
+ | == Name == | ||
+ | This theorem has been widely known for a while now. The AoPS user solasky rediscovered it and put a name to it. You can call it whatever you want. |
Latest revision as of 01:10, 27 September 2021
Theorem
Given nonnegative real numbers and that
is fixed and all the terms inside the sum are nonnegative, the maximum value of
is when
Proof
The weighted AM-GM Inequality states that if are nonnegative real numbers, and
are nonnegative real numbers (the "weights") which sum to 1, then
We let
and
in this inequality. We get that
Dividing both sides and then taking to the nth power, we get
Then we can multiply both sides to get
The equality case for the weighted AM-GM inequality is when all the
terms such that
is not
are equal, or in this case
Example Problems
Problem 1
Given nonnegative integers such that
find the maximum value
Solution 1
By the new theorem, we know that so
Plugging it in, our answer is
Name
This theorem has been widely known for a while now. The AoPS user solasky rediscovered it and put a name to it. You can call it whatever you want.