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I recently came across a question regarding quadratic equations. I encountered this question in a maths Olympiad based book.

Find all the positive integers n such that the equation

$a_{n+1}$$x^2$ -$2$$x$$\sqrt{\sum_{i=1}^{n+1}a_{i}^2}$ +$\sum_{i=1}^na_{i}$=0

has real roots for every choice of real numbers $a_{1},a_{2}....a_{n+1}$

My initial approach was to take the discriminant positive... But that didn't help much. Might be possible, I lack proper information. I didn't understand what is so special about $a_{n+1}$. Could have been any coefficient.Can any inequality be applied? I also tried to apply the inequality that root mean square of$a_1,a_2,....a_{n}$ is greater than the mean $a_1,a_2,....a_{n}$... Please point me in the right direction. Any help would be appreciated.

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    $\begingroup$ I have tried to edit the question... Please have a look again $\endgroup$ – user440009 May 12 '17 at 12:32
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Let $ \Delta $ be the discriminant of the equation $$ \Delta=4\sum_{i=1}^{n+1}a_{i}^{2}-4a_{n+1}\sum_{i=1}^{n}a_{i} $$ We must have that $ \Delta \geq 0 $ which implies that $$ \sum_{i=1}^{n+1}a_{i}^{2} \geq a_{n+1}\sum_{i=1}^{n}a_{i} $$ Rewrite the last inequality as $$ a_{n+1}^{2}-a_{n+1}\sum_{i=1}^{n}a_{i}+\sum_{i=1}^{n}a_{i}^{2} \geq 0 $$ Since this must hold for any $ a_{n+1} \in \mathbb{R} $, we must have that the new discriminant $$\Delta '=\big(\sum_{i=1}^{n}a_{i}\big)^{2}-4\sum_{i=1}^{n}a_{i}^{2} \leq 0 $$ And the last inequality must hold for any real numbers $ a_{1}, \dots , a_{n} $. We show that only $ n \in \{ 1,2,3,4\} $ satisfy this property.

So let $ n \geq 5 $and consider the example $ a_{i}=i $ $ \forall i \in \overline{1,n} $ . Then in this case, we must have that $$\Delta '=\big(\sum_{i=1}^{n}i\big)^{2}-4\sum_{i=1}^{n}i^{2}=\frac{n^{2}(n+1)^{2}}{4}-4\frac{n(n+1)(2n+1)}{6} \leq 0 $$ Dividing the last inequality by $ \frac{n(n+1)}{12} $, we see that we must have $$ 3n(n+1) -8(2n+1) \le 0 $$

Computation then shows that $$ n \in \big[\frac{13-\sqrt{265}}{6},\frac{13+\sqrt{265}}{6}\big] $$

But $ \sqrt{265}<17 $ so $ n<5 $. Therefore the only remaining possibilities are $ n \in \{1,2,3,4\} $ and it is easily verified that all of them actually occur.

Indeed, for $ n=1 $, we must have that $ a_{1}^{2}-4a_{1}^{2} \leq 0 $ for any real number $ a_{1} $ which clearly holds.

For $ n=2$, we must have that $$ (a_{1}+a_{2})^{2}-4(a_{1}^{2}+a_{2}^{2})\leq 0 $$ which can be rewritten as $$ -(a_{1}-a_{2})^{2}-2(a_{1}^{2}+a_{2}^{2}) \leq 0 $$ which again holds true for any real numbers $ a_{1} $ and $ a_{2} $.

For $ n=3 $ we must have that $$(a_{1}+a_{2}+a_{3})^{2}-4(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})\leq 0 $$ but by Cauchy-Scwarz, we have that $$(a_{1}+a_{2}+a_{3})^{2} \leq 3(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}) $$ so clearly the inequality needed for $ n=3 $ holds as well.

Finally, for $n=4 $ what we need is exactly the Cauchy-Schwarz inequality $$(a_{1}+a_{2}+a_{3}+a_{4})^{2} \leq 4(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}) $$

We conclude that $ n \in \{1,2,3,4\} $.

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  • $\begingroup$ You're welcome! I'm glad I could help. $\endgroup$ – Raizen May 12 '17 at 17:35

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