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Show that if $n ∈ \mathbb{N}$ and $a_1,...,a_n$ are nonnegative real numbers, then $(a_1+···+a_n)^2 ≤ n(a^2_1+···+a^2_n).$

What I have tried:

Set $t = a_1+ ...+ a_{n-1}$

Then square the LHS as so $$(t+a_n)^2 =t^2+2ta_n+a^2_n$$ $$=t^2+ta_n+ta_n+a^2_n$$ $$=t(t+a_n)+a_n(t+a_n)$$ $$=(t+a_n)(t+a_n)$$ $$\implies (t+a_n)(t+a_n) \le n(t^2+a_n^2)$$ $$\implies(t+a_n)\le n(t+a_n)$$

Which would imply that for all $n \in \mathbb{N}$ the equality holds. Does my proof work?

me.limes
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  • I don't get your line of reasoning? How from $(t+a_n)^2 = (t+a_n)(t+a_n)$ (which by the way is easier to establish) you get $(t+a_n)(t+a_n) \le n(t^2+a_n^2)$? But your problem can be easily proved by Cauchy-Schwarz inequality. – Presage Nov 21 '21 at 21:11
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    Let me follow up on Dominik's comment: Let $a=(a_1,\ldots,a_n), b=(1,\ldots,1)\in\Bbb R^n.$ Then $|\langle a,b\rangle^2\le|b|^2|a|^2.$ I don't think you have proven anything. Do you want to prove it by induction? – PinkyWay Nov 21 '21 at 21:15
  • You can also use the convexity of the function $f(x)=x^2$ and apply Jensen. – PinkyWay Nov 21 '21 at 21:23
  • Since $a_1,\ldots,a_n\ge 0,$ you can use the QM-AM inequality. – PinkyWay Nov 21 '21 at 21:26
  • @Invisible Is that inequality named the Cauchy-Schwarz inequality? Seems like that's exactly what I need. Thanks – me.limes Nov 21 '21 at 21:27
  • No, this does not work at all. The consequent of the first implication is the thesis, and does not follow from the trivial $(t+a_n)^2=(t+a_n)(t+a_n)$. –  Nov 21 '21 at 21:33
  • @me.limes, you should study systematically. This question belongs to either linear algebra or algebra-precalculus and, seeing your questions, you even do ODE's and probability. First learn the theory step by step and then search the site. (=: – PinkyWay Nov 21 '21 at 21:35
  • Divide both sides by $n^2$. Then you see that $\bar{a}^2\leq\bar{a^2}$. Since the squaring operation is a convex function, we see that the inequality is true (by Jensen's inequality). – lmaosome Nov 21 '21 at 21:36

1 Answers1

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Want to show that $(\sum_{k=1}^n a_k)^2 \le n\sum_{k=1}^n a_k^2 $ or $s_1(n)^2 \le n s_2(n) $ where $s_m(n) =\sum_{k=1}^n a_k^m $.

True for $n=1$ (trivial) and $n=2$ (easy).

If true for $n$, then

$\begin{array}\\ s_1^2(n+1) &=(s_1(n)+a_{n+1})^2\\ &=s_1^2(n)+2a_{n+1}s_1(n)+a_{n+1}^2\\ &\le ns_2(n)+2a_{n+1}\sqrt{n s_2(n)}+a_{n+1}^2\\ \end{array} $

so we want

$\begin{array}\\ ns_2(n)+2a_{n+1}\sqrt{n s_2(n)}+a_{n+1}^2 &\le (n+1)s_2(n+1)\\ &= (n+1)(s_2(n)+a_{n+1}^2)\\ &= (n+1)s_2(n)+(n+1)a_{n+1}^2\\ \end{array} $

or $2a_{n+1}\sqrt{n s_2(n)} \le s_2(n)+na_{n+1}^2 $

or

$\begin{array}\\ 0 &\le na_{n+1}^2-2a_{n+1}\sqrt{n s_2(n)}+s_2(n)\\ &=n\left(a_{n+1}^2-2a_{n+1}\sqrt{ \dfrac{s_2(n)}{n}}+\dfrac{s_2(n)}{n}\right)\\ &=n\left(a_{n+1}-\sqrt{ \dfrac{s_2(n)}{n}}\right)^2\\ \end{array} $

which is true with equality iff $a_{n+1}=\sqrt{ \dfrac{s_2(n)}{n}} $.

marty cohen
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