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(NOTE: Already posted here, but closed without an answer)

Hi, I've been trying to complete the following question:

Suppose we have two vectors of $n$ real numbers, $[x_1,x_2,⋯,x_n]$ and $[y_1,y_2 ⋯,y_n]$ and the following inequality holds: $$(x_1y_1+x_2y_2+⋯+x_ny_n)^2≤(x_1^2+x_2^2+⋯+x_n^2)(y_1^2+y_2^2+⋯+y_n^2)$$ Show that if the sum of components of one vector adds up to 1 then the sum of the squares of the same vector is at least $\frac 1n$.

I've tried a few avenues, and have come up a proof that I am not confident is right.

Proof by induction:

Base case is $n=1$, which is trivial, since $x_1^2 = 1^2 = 1$ and so $1 \ge \frac 1 1$. Therefore base case is true.

Assume it is true for n.

$$x_1^2+...+x_n^2+x_{n+1}^2 \ge \frac 1 {n+1}$$

Since $x_1^2+...+x_n^2 \ge \frac 1 n$ by our assumption,

$$\frac 1 n + x_{n+1}^2 \ge \frac 1 {n+1}$$ It is this step that I think is incorrect, as the $x_1^2+...+x_n^2$ must get smaller in order to accomodate the new value of $x_{n+1}$, and still remain equal to 1. Therefore I don't think I can do this step?

$$x_{n+1}^2 \ge \frac 1 {n+1} - \frac 1 n$$

The left hand side must always be $\ge 0$ and the right hand side must always be negative for values of $n \ge 1$. Therefore true for $n+1$, so must be true for all $n$. QED.

Can you confirm it is wrong? If it is wrong, could you please explain how to prove it in your answer. Thanks.

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    $+1$ for an excellent question. Have you tried to make sense of J-Moravitz's comment in the post you have linked? If so, what part of it doesn't make sense, because that literally has the answer in it. – 2017-01-12
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    Your intuition correct, the induction argument isn't correct. There are 2 components to the theorem: the set being generalized over, and the claim made of the set. You can either characterize it as: (a) the set being generalized over is $\mathbb N$, the theorem is $\sum x = 1 \implies \text{inequality}$, or (b) the set generalized over is all vectors with $\sum x = 1$ and the theorem is $\text{inequality}$. Your proof is attempting to use the set from from $(a)$ and the theorem from $(b)$. Either is a correct, but mixed up isn't correct. – 2017-01-12
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    @астон вілла олоф мэллбэрг Please see my updated question with my new proof attempt. It took me awhile because I had to teach myself the basic vector definitions before I could see why the Cauchy-Schwarz was useful. Is my proof correct and clear? – 2017-01-12
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    I think it is fine now! – 2017-01-12
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    Excellent, thanks for your help – 2017-01-12

2 Answers 2

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You can prove it using Jensen's inequality for $f(x) = x^2$. Since $x^2$ opens upwards,

$$f(\text{avg}~ a) \le \text{avg}~f(a)$$

More specifically:

$$\left(\frac 1n \sum_{k=1}^n a_k\right)^2 \le \frac 1n \left(\sum_{k=1}^n a_k{}^2\right)$$

The rest is just algebra, use $\sum_{k=1}^n a_k = 1$ to establish $\sum_{k=1}^n a_k{}^2 \ge \frac{1}{n}$.

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    open upwards = convex – 2017-01-12
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    I can only keep concave/convex straight when dealing with shapes. Concave means the shape has a cave in it. – 2017-01-12
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    A graph that looks like a "V" is conVex. – 2017-01-12
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Hint: Try to look at the inequality with $y_k=1,\forall i=1,...,n$