# Prove that if sum of components of one vector adds up to 1 then the sum of the squares of the same vector is at least 1/n

This is regarding the proof for another question.

# Problem statement:

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 $\frac 1n$.

# Proof given:

Assuming the following inequality holds: $$(x_1y_1+x_2y_2+⋯+x_ny_n)^2≤(x_1^2+x_2^2+⋯+x_n^2)(y_1^2+y_2^2+⋯+y_n^2)$$

Set each component in the y vector to 1. This gives:

$$(x_1y_1+x_2y_2+⋯+x_ny_n)^2≤n(x_1^2+x_2^2+⋯+x_n^2)$$

Dividing by n:

$$\frac {(x_1y_1+x_2y_2+⋯+x_ny_n)^2} n ≤(x_1^2+x_2^2+⋯+x_n^2)$$

As $x_1y_1+x_2y_2+⋯+x_ny_n = 1$:

$$\frac 1 n \le x_1^2+x_2^2+⋯+x_n^2$$ QED.

# My questions:

1. I don't get the Set each component in the y vector to 1 step. Why is this assumption made?
• You have that $y_i=1\forall i$, then $\sum x_iy_i=\sum x_i$. You dont need to carry the $y_i$ terms. If you made the assumption then your inequality becomes $$\sum x_i\le n\sum x_i^2$$ – Masacroso Jan 24 '17 at 19:09
• Thanks @Masacroso . But I still don't get how ∑yi^2 = 1 – bcp Jan 24 '17 at 19:14
• We have that $\sum y_i^2=n$, because $y_i^2=1^2=1$ for all $i$. – Masacroso Jan 24 '17 at 19:15

Your concerns are adequate, at least if the formulation of the given statement is exactly as that:

"Suppose we have two vectors of $n$ real numbers, $[x_1,x_2,⋯,x_n]$ and $[y_1,y_2.⋯,y_n]$ ..."

means that we may not use the stated inequality for any other vectors. Therefore, picking the $y_i$ as we please is not logically sound.

However, the inequality you are given actually holds for all vectors, and that does allow us to pick the $y_i$.

• Alright... I get it. Thanks. – bcp Jan 24 '17 at 19:18

Given: We have two vectors [x1, x2, x3, …. xn] and [y1, y2, y3, …… yn]. Since the inequality holds true for all real numbers, we may start with the identity value, 1, considering any of the two vector. This will make it easier to achieve the proof. Let us say, all components of vector x are 1. Then the inequality holds the form of: (y1+y2+y3+……..+yn)2 < (1+1+1+1……..+1)(y12+y22+y32+…..+yn2)

As per the condition given, when sum of the components add up to 1, then y1+y2+y3+……..+yn = 1 Substituting the value, we have: 1 < n. (y12+y22+y32+…..+yn2) 1/n < (y12+y22+y32+…..+yn2) Q.E.D