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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?

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    $\begingroup$ You need to restrict the domain. It's false for $a=1, b=-1$ for example. $\endgroup$
    – lulu
    Commented Nov 9, 2020 at 22:33
  • $\begingroup$ @lulu sorry, I just found the issue, fixed it. Can you look into the question again. $\endgroup$ Commented Nov 9, 2020 at 22:39
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    $\begingroup$ Looks inductiony. $a^2+b^2+c^2 < (a+b)^2 + c^2 < (a+b+c)^2$ $\endgroup$ Commented Nov 9, 2020 at 22:40
  • $\begingroup$ But if all the terms are $0$ the problem is trivial. Just expand the expression on the right and cancel matcheing terms on the left. What;'s left is strictly positive. $\endgroup$
    – lulu
    Commented Nov 9, 2020 at 22:40

2 Answers 2

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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...

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  • $\begingroup$ Thanks. Though I have accepted the other answer, this one is more intuitive. +1 $\endgroup$ Commented Nov 9, 2020 at 22:50
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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}

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