Proving $a^2+b^2+c^2+\cdots+z^2<(a+b+c+\cdots+z)^2$

Question

I know and can prove that, $a^2 + b^2 < (a+b)^2$ where $a > 0$ and $b > 0$

Now, how can I prove the generalization, i.e.

$$ a^2 + b^2+ c^2 + \cdots + z^2 < (a +b + c+ \cdots + z)^2 $$

What is the way to reach that conclusion from the above statement?

Answer 1

Induct. For instance, \begin{align*} a^2 + b^2 + c^2 &= (a^2 + b^2) + c^2 \\ &< (a+b)^2 + c^2 & & \text{base case on $a$, $b$}\\ &< ((a+b)+c)^2 & & \text{base case on $a+b$, $c$} \\ &= (a+b+c)^2 \text{.} \end{align*}

You should be able to see how to adapt this to a generic induction step...

Answer 2

Let $a_1,...,a_n\geq 0$. Then\begin{align} (a_1+\cdots+a_n)^2&=\left(\sum_{i=1}^na_i\right)^2\\ &=\sum_{i=1}^n\sum_{j=1}^na_ia_j\\ &=a_1^2+\cdots+a_n^2+\sum_{i\neq j}a_ia_j\\ &\geq a_1^2+\cdots+a_n^2. \end{align}