# The sum of the squares is less than or equal to the square of the sums for all $n$

I am trying to understand this proof. Rather an important part of the proof. I have already shown this is true for $$n=2$$ and am assuming the $$a_n$$ case is true.

$$(a_1^2 +a_2^2 +\ldots +a_n^2) \le (a_1 +a_2 +\ldots +a_n)^2$$ Want to show that \begin{align} (a_1^2 +a_2^2 +\ldots +a_n^2 +a_{n+1}^2) & \le (a_1 +a_2 +\ldots +a_n +a_{n+1})^2 \\ &= (a_1^2 +a_2^2 +\ldots +a_n^2) +a_{n+1}^2 \\ & \le ((a_1 +a_2 +\ldots +a_n) +(a_{n+1}))^2 \\ & = (a_1^2 +a_2^2 +\ldots +a_n^2 + a_{n+1}^2) \\ & \le (a_1 +a_2 +\ldots +a_n)^2 +2(a_1 +a_2 +\ldots +a_n)(a_{n+1}) +(a_{n+1})^2 \end{align} and here is the part I am not understanding. For some reason the proof moves some of the terms over and I cannot identify what is being replaced or why. My guess is that the terms that moves are the $${n+1}$$ terms. But, I am not certain.

$$a_1^2+a_n^2+a_{n+1}^2+\ldots+2(a_1+a_2+\ldots+a_n)(a_{n+1}) \le (a_1+a_2+\ldots+a_n)^2$$

• Your first line does not imply your second. The second line is what you need to show. I think that if you were to write it out carefully you might be able to figure it out on your own. Commented Mar 8, 2015 at 5:59
• please read this about induction to see how to structure an inductive proof. In this case if all the $a_i$ are positive a direct proof is easier. If you expand the right side you get a bunch of terms that match the left and some more as well. Subtract the matching ones and you are done. Commented Mar 8, 2015 at 6:00
• Try using binomial expansion on the last term of eq. 3.
– BHT
Commented Mar 8, 2015 at 6:04
• Thank you @Ross and Ethan. Commented Mar 8, 2015 at 6:51

## 2 Answers

If you really need to use induction, here's what you need for the inductive step:

Assuming $(a_1^2+a_2^2+...+a_n^2 ) \le (a_1+a_2+...+a_n)^2$

then \begin{align} (a_1+a_2+...+a_n+a_{n+1})^2 &= ((a_1+a_2+...a_n)+(a_{n+1}))^2 \\ &= (a_1+a_2+...+a_n)^2+2(a_1+a_2+...+a_n)a_{n+1} + a_{n+1}^2 \\ &\ge (a_1+a_2+...+a_n)^2 + a_{n+1}^2 \tag{*}\\ &\ge (a_1^2+a_2^2+...+a_n^2 ) + a_{n+1}^2 \\ \end{align}as required. Note that $2(a_1+a_2+...+a_n)a_{n+1}\ge 0)$ for the (*) step.

You do require that all the $a_i$ are not negative.

Being visual myself, I prefer pictures....

• I always knew this from induction, but this picture is worth 1000 proof steps :)
– John
Commented Nov 28, 2020 at 18:29

inductive step:

the claim being correct for $$(a_1^2+a_2^2+...+a_n^2) \le (a_1+a_2+...+a_n)^2$$ implies $$(a_1^2+a_2^2+...+a_n^2+a_{n+1}^2) \le (a_1+a_2+...+a_n+a_{n+1})^2$$ Proof \begin{align} a_1^2+a_2^2+...+a_n^2+a_{n+1}^2 &=(a_1^2+a_2^2+...+a_n^2)+a_{n+1}^2\\ & \leq (a_1+a_2+...+a_n)^2+a_{n+1}^2 \mbox{ (using the assumption)}\\ &=y^2+a_{n+1}^2 \mbox{ (rewriting } y=a_1+a_2+...+a_n)\\ & \leq (y+a_{n+1})^2 \mbox{ (using: } a^2+b^2\leq (a+b)^2)\\ &=(a_1+a_2+...+a_n+a_{n+1})^2 \mbox{ (plug back for } y) \end{align}

• This proof seems much better than the chosen answer since it does not require assuming that anything is positive. Would you agree? Commented Jun 17, 2017 at 18:11
• @EternusVia - you need to have the $a_i$ positive for the statement to be assuredly true. Consider $2^2+(-2)^2>(2+(-2))^2$. Commented Nov 22, 2017 at 0:32
• @Joffan you are of course right. Commented Nov 23, 2017 at 8:41