Prove a^2_1+3a^2_2+5a^2_3+....+(2n−1)a^2_n≤1

[lfdahl]

- except for this one: :) Given: $a_1 \ge a_2 \ge .. \ge a_n \ge 0$ and the constraint: $\sum_{i=1}^{n}a_i = 1$.

Prove, that

\[a_1^2+3a_2^2+5a_3^2+...+(2n-1)a_n^2 \le 1\]

 

[Albert1]

- except for this one: :) Given: $a_1 \ge a_2 \ge .. \ge a_n \ge 0$ and the constraint: $\sum_{i=1}^{n}a_i = 1$.

Prove, that

\[a_1^2+3a_2^2+5a_3^2+...+(2n-1)a_n^2 \le 1\]

my solution:

Spoiler

using $AP\geq GP$
let $A=a_1^2+3a_2^2+5a_3^2+...+(2n-1)a_n^2$
for $a_1+a_2+------+a_n=1\geq n(\sqrt [n]{a_1a_2----a_n})=B$
we have :$max(B)$ will occur at $a_1=a_2=----=a_n=\dfrac{1}{n}$
so $max(A)\leq (1+3+5+----+(2n-1))\times \dfrac{1}{n^2}$
$=\dfrac {2n^2}{2n^2}=1$

 

[lfdahl]

my solution:

Spoiler

using $AP\geq GP$
let $A=a_1^2+3a_2^2+5a_3^2+...+(2n-1)a_n^2$
for $a_1+a_2+------+a_n=1\geq n(\sqrt [n]{a_1a_2----a_n})=B$
we have :$max(B)$ will occur at $a_1=a_2=----=a_n=\dfrac{1}{n}$
or at $(a_1=1,and \,\, a_2=a_3=--------=a_n=0)$
so $max(A)\leq (1+3+5+----+(2n-1))\times \dfrac{1}{n^2}$
$=\dfrac {2n^2}{2n^2}=1$

Hello, Albert!

Spoiler

I have one question to your solution:
If I let $f(a_1,a_2,...,a_n) = a_1^2+3a_2^2+5a_3^2+...+(2n-1)a_n^2$, does the following implication hold:
$B_{max} \Rightarrow f_{max}$?

The reason, I ask, is, that I have solved the problem by means of the Lagrange multiplier, and it turns out, that in the critical point I find descending (different) $a$-values...: $a_1 > a_2 > ...> a_n$.

 

[Albert1]

Hello, Albert!

Spoiler

I have one question to your solution:
If I let $f(a_1,a_2,...,a_n) = a_1^2+3a_2^2+5a_3^2+...+(2n-1)a_n^2$, does the following implication hold:
$B_{max} \Rightarrow f_{max}$?

The reason, I ask, is, that I have solved the problem by means of the Lagrange multiplier, and it turns out, that in the critical point I find descending (different) $a$-values...: $a_1 > a_2 > ...> a_n$.

Spoiler

$B_{max} \Rightarrow f_{max}$? yes
but $f_{max} \Rightarrow B_{max}$ no
for $ B_{max}$ occurs at $a_1=a_2=----=a_n=\dfrac {1}{n}$
but $f_{max}$ will occur at many points
just name a few:
$(a_1,a_2,a_3,------a_n)=(1,0,0,-----,0)$
$(a_1,a_2,a_3,------a_n)=(0.5,0.5,0,-----,0)$
$(a_1,a_2,a_3,------a_n)=(\dfrac{1}{3},\dfrac {1}{3},\dfrac {1}{3},-----,0)$
--------
$(a_1,a_2,a_3,------a_n)=(\dfrac{1}{n},\dfrac{1}{n},\dfrac{1}{n},---,\dfrac{1}{n})$
but all those points will have the same values of $f_{max}=1$
It is much more complicated solving the problem by means of the Lagrange multiplier,
how do you think ?

 

[lfdahl]

Spoiler

$B_{max} \Rightarrow f_{max}$? yes
but $f_{max} \Rightarrow B_{max}$ no
for $ B_{max}$ occurs at $a_1=a_2=----=a_n=\dfrac {1}{n}$
but $f_{max}$ will occur at many points
just name a few:
$(a_1,a_2,a_3,------a_n)=(1,0,0,-----,0)$
$(a_1,a_2,a_3,------a_n)=(0.5,0.5,0,-----,0)$
$(a_1,a_2,a_3,------a_n)=(\dfrac{1}{3},\dfrac {1}{3},\dfrac {1}{3},-----,0)$
--------
$(a_1,a_2,a_3,------a_n)=(\dfrac{1}{n},\dfrac{1}{n},\dfrac{1}{n},---,\dfrac{1}{n})$
but all those points will have the same values of $f_{max}=1$
It is much more complicated solving the problem by means of the Lagrange multiplier,
how do you think ?

I think, you´re right! Thanks a lot for your participation and correct solution!

 

[Albert1]

I think, you´re right! Thanks a lot for your participation and correct solution!

Spoiler

you are also right ,the benefit of Lagrange multiplier is :
we can find a solution $a_1>a_2>a_3>------->a_n$
given $a_1+a_2+a_3+-----+a_n=1---(1)$
$a_n=\dfrac {k}{4n-2},k>0$ and will satisfy (1)
in this case $f(max)=a_1^2+3a_2^2+-----+(2n-1)a_n^2<1$

 

[lfdahl]

Spoiler

you are also right ,the benefit of Lagrange multiplier is :
we can find a solution $a_1>a_2>a_3>------->a_n$
given $a_1+a_2+a_3+-----+a_n=1---(1)$
$a_n=\dfrac {k}{4n-2},k>0$ and will satisfy (1)
in this case $f(max)=a_1^2+3a_2^2+-----+(2n-1)a_n^2<1$

Hello, again, Albert!:D

Below is my solution:

Spoiler

I think your $k$ is the same as the Lagrange multiplier ($\lambda$).

With \[f(a_1,a_2,...,a_n) = \sum_{j=1}^{n}(2j-1)a_j^2,\: \: \: g(a_1,a_2,...,a_n) = \sum_{i=1}^{n}a_i = 1\]
- I get:
\[\nabla f = 2\left ( a_1,3a_2,5a_3,...,(2n-1)a_n \right ), \: \: \: \lambda \nabla g = \lambda \underbrace{(1,1,..,1)}_{n\: \: times}\\\\ \nabla f = \lambda \nabla g \Rightarrow a_j^* = \frac{\lambda}{2} \cdot \frac{1}{2j-1}, \: \: \: j = 1,2,...n\]
Determining $\lambda$ from the constraint:
\[\frac{\lambda}{2} \sum_{j=1}^{n} \frac{1}{2j-1} = 1 \rightarrow \lambda = \frac{2}{\sigma},\: \: \: \sigma = \sum_{j=1}^{n} \frac{1}{2j-1} \geq \frac{n^2}{\sum_{j=1}^{n}(2j-1)} = \frac{n^2}{n^2}=1.\]
The inequality is a consequence of the harmonic-mean geometric-mean arithmetic-mean inequality:
\[\frac{n}{\sum_{i=1}^{n}\frac{1}{x_i}} \leq \left ( \prod_{i=1}^{n}x_i \right )^{1/n} \leq \frac{\sum_{i=1}^{n}x_i}{n}\]
Determining $f$´s value in the critical point:
\[f(a_1^*,a_2^*,...,a_n^*) = \sum_{j=1}^{n}(2j-1)(a_j^*)^2 \\\\ = \sum_{j=1}^{n}(2j-1)\left ( \frac{\lambda }{2} \right )^2 \frac{1}{(2j-1)^2} = \left ( \frac{\lambda }{2} \right )^2 \sigma = \frac{1}{\sigma }.\]
But $\sigma \ge 1$, and we´re done.

 

[Albert1]

Hello, again, Albert!:D

Below is my solution:

Spoiler

I think your $k$ is the same as the Lagrange multiplier ($\lambda$).

With \[f(a_1,a_2,...,a_n) = \sum_{j=1}^{n}(2j-1)a_j^2,\: \: \: g(a_1,a_2,...,a_n) = \sum_{i=1}^{n}a_i = 1\]
- I get:
\[\nabla f = 2\left ( a_1,3a_2,5a_3,...,(2n-1)a_n \right ), \: \: \: \lambda \nabla g = \lambda \underbrace{(1,1,..,1)}_{n\: \: times}\\\\ \nabla f = \lambda \nabla g \Rightarrow a_j^* = \frac{\lambda}{2} \cdot \frac{1}{2j-1}, \: \: \: j = 1,2,...n\]
Determining $\lambda$ from the constraint:
\[\frac{\lambda}{2} \sum_{j=1}^{n} \frac{1}{2j-1} = 1 \rightarrow \lambda = \frac{2}{\sigma},\: \: \: \sigma = \sum_{j=1}^{n} \frac{1}{2j-1} \geq \frac{n^2}{\sum_{j=1}^{n}(2j-1)} = \frac{n^2}{n^2}=1.\]
The inequality is a consequence of the harmonic-mean geometric-mean arithmetic-mean inequality:
\[\frac{n}{\sum_{i=1}^{n}\frac{1}{x_i}} \leq \left ( \prod_{i=1}^{n}x_i \right )^{1/n} \leq \frac{\sum_{i=1}^{n}x_i}{n}\]
Determining $f$´s value in the critical point:
\[f(a_1^*,a_2^*,...,a_n^*) = \sum_{j=1}^{n}(2j-1)(a_j^*)^2 \\\\ = \sum_{j=1}^{n}(2j-1)\left ( \frac{\lambda }{2} \right )^2 \frac{1}{(2j-1)^2} = \left ( \frac{\lambda }{2} \right )^2 \sigma = \frac{1}{\sigma }.\]
But $\sigma \ge 1$, and we´re done.

Spoiler

take $a_1+a_2=1$ ,for simplicity
using your method we will find
$max(a_1^2+3a_2^2-\lambda(a_1+a_2-1))$
in this case :$a_1=\dfrac{\lambda}{2}$ ,$a_2=\dfrac{\lambda}{6}$
and $\lambda=\dfrac {3}{2}$ (fixed)
so $a_1=\dfrac{3}{4}$,$a_2=\dfrac{1}{4}$
and $a_1^2+3a_2^2=\dfrac{12}{16}=\dfrac {3}{4}<1$

 

[lfdahl]

I can´t help it to show the short and elegant solution below:

Spoiler

\[1 = \left ( \sum_{i=1}^{n}a_i \right )^2 = \sum_{i=1}^{n}a_i^2+2 \sum_{1\leq i < j \leq n}a_ia_j \geq \sum_{i=1}^{n}a_i^2 + 2\sum_{j=2}^{n}(j-1)a_j^2 = \sum_{i=1}^{n}(2i-1)a_i^2\]