Difference between revisions of "SFMT"

Revision as of 15:33, 19 September 2021

Theorem

Given nonnegative real numbers $e_1, e_2, e_3, \cdots ,e_n$ and that $x_1+x_2+x_3+\cdots+x_n$ is fixed and all the terms inside the sum are nonnegative, the maximum value of $x_1^{e_1}x_2^{e_2}x_3^{e_3}\cdots x_n^{e_n}$ is when $\dfrac{x_1}{e_1}=\dfrac{x_2}{e_2}=\dfrac{x_3}{e_3}=\cdots =\dfrac{x_n}{e_n}.$

Proof

The weighted AM-GM Inequality states that if $a_1, a_2, \dotsc, a_n$ are nonnegative real numbers, and $\lambda_1, \lambda_2, \dotsc, \lambda_n$ are nonnegative real numbers (the "weights") which sum to 1, then $\lambda_1 a_1 + \lambda_2 a_2 + \dotsb + \lambda_n a_n \ge a_1^{\lambda_1} a_2^{\lambda_2} \dotsm a_n^{\lambda_n}.$ We let $\lambda_i=\dfrac{e_i}{e_1+e_2+\cdots+e_n}$ and $a_i=\dfrac{x_i}{e_i}$ in this inequality. We get that $x_1+x_2+x_3+\cdots+x_n \ge \sqrt[n]{\dfrac{x_1^{e_1}x_2^{e_2}x_3^{e_3}\cdots x_n^{e_n}}{e_1^{e_1}e_2^{e_2}e_3^{e_3}\cdots e_n^{e_n}}}(e_1+e_2+e_3+\cdots+e_n).$ Dividing both sides and then taking to the nth power, we get $\left(\dfrac{x_1+x_2+x_3+\cdots+x_n}{e_1+e_2+e_3+\cdots+e_n}\right)^n \ge \dfrac{x_1^{e_1}x_2^{e_2}x_3^{e_3}\cdots x_n^{e_n}}{e_1^{e_1}e_2^{e_2}e_3^{e_3}\cdots e_n^{e_n}}.$ Then we can multiply both sides to get $x_1^{e_1}x_2^{e_2}x_3^{e_3}\cdots x_n^{e_n} \le \left(\dfrac{x_1+x_2+x_3+\cdots+x_n}{e_1+e_2+e_3+\cdots+e_n}\right)^n (e_1^{e_1}e_2^{e_2}e_3^{e_3}\cdots e_n^{e_n}).$ The equality case for the weighted AM-GM inequality is when all the $a_i$ terms such that $e_i$ is not $0$ are equal, or in this case $\dfrac{x_1}{e_1}=\dfrac{x_2}{e_2}=\dfrac{x_3}{e_3}=\cdots=\dfrac{x_n}{e_n}.$

Example Problems

Problem 1

Given nonnegative integer x, y, z such that $x+y+z=16,$ find the maximum value $x^3y^2z^3.$

Solution 1

By the new theorem, we know that $\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{3},$ so x=6, y=4, z=6. Plugging it in, our answer is $746496.$

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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>
 == Proof ==
-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>[/quote] 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>
+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 ===

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