(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.
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 $\sum_{k=1}^n a_k{}^2 \ge \frac{1}{n}$.