https://artofproblemsolving.com/community/c6h368122_prove_that_the_least_of_the_numbers_does_not_exceed_110

From the OMC RMO livesolve. solved with rawlat_vanak, AdhityaMV, zvfrozel, and mathscrazy. Supposing otherwise, we have $|a_i-a_j| > \frac{1}{\sqrt{5}}$ for all $i \ne j$. We may order $a_1 \leq \dots \leq a_5$. Now for any $i < j$ we have \[(a_j-a_i) > \frac{(j-i)}{\sqrt{5}}\]and thus \[ 4 \geq 4 \left( \sum_{i} a_i^2 \right) - \left( \sum_{i} a_i \right)^2 = \sum_{i < j} (a_i - a_j)^2 > \sum_{i < j} \frac{1}{5} (j - i)^2 = 4 \]a contradiction. $\square$

Redacted Thanks for pointing out the error

SomeonecoolLovesMaths wrote: Wolog, let $a_1 < a_2 < a_3 < a_4$. If possible let, $|a_i - a_j| > \frac{1}{\sqrt{5}} \; \forall \; i,j \in \{1,2,3,4\} \; \; i \neq j$. Then, $$1 = a_1^2 + a_2^2 + a_3^2 + a_4^2 > a_1^2 + \left( \frac{1}{\sqrt{5}} + a_1 \right)^2 + \left( \frac{2}{\sqrt{5}} + a_1 \right)^2 + \left( \frac{3}{\sqrt{5}} + a_1 \right)^2$$$$\Longrightarrow 4a_1^2 - \left( \frac{12}{\sqrt{5}} \right) a_i + \frac{14}{5} < 1$$$$\Longrightarrow 4a_1^2 - \left( \frac{12}{\sqrt{5}} \right) a_i + \frac{9}{5} = 4 \left( a_i - \frac{3}{2 \sqrt{5}} \right)^2 < 0$$Impossible! Thus $\exists$ $i,j$ with $ 1 \leq i < j \leq 4$, such that $(a_i - a_j)^2 \leq \frac{1}{5}$. I am surprised they put an ineq like this, also for the love of god I can't remember where i hv seen this before.

We will use contradiction. Suppose otherwise. By WLOG, suppose that $a_1\le a_2\le a_3\le a_4$. The inverse of the condition rewrites as $$|a_i-a_j|> \frac{1}{\sqrt{5}}\implies a_j-a_i>\frac{j-i}{\sqrt{5}}$$for $j>i$. Note that $$4 \ge 4(a_1^2+a_2^2+a_3^2+a_4^2)-(a_1+a_2+a_3+a_4)^2$$$$=3(a_1^2+a_2^2+a_3^2+a_4^2)-(\sum_{sym} a_1a_2)$$$$=\sum_{i < j} (a_i-a_j)^2$$$$>\frac{1}{5}\sum_{i<j}(j-i)^2$$$$=\frac{9+4+4+1+1+1}{5}=4,$$a contradiction.

SomeonecoolLovesMaths wrote: Wolog, let $a_1 < a_2 < a_3 < a_4$. If possible let, $|a_i - a_j| > \frac{1}{\sqrt{5}} \; \forall \; i,j \in \{1,2,3,4\} \; \; i \neq j$. Then, $$1 = a_1^2 + a_2^2 + a_3^2 + a_4^2 > a_1^2 + \left( \frac{1}{\sqrt{5}} + a_1 \right)^2 + \left( \frac{2}{\sqrt{5}} + a_1 \right)^2 + \left( \frac{3}{\sqrt{5}} + a_1 \right)^2$$$$\Longrightarrow 4a_1^2 - \left( \frac{12}{\sqrt{5}} \right) a_i + \frac{14}{5} < 1$$$$\Longrightarrow 4a_1^2 - \left( \frac{12}{\sqrt{5}} \right) a_i + \frac{9}{5} = 4 \left( a_i - \frac{3}{2 \sqrt{5}} \right)^2 < 0$$Impossible! Thus $\exists$ $i,j$ with $ 1 \leq i < j \leq 4$, such that $(a_i - a_j)^2 \leq \frac{1}{5}$. Isn't this a fakesolve ? cuz x > y doesn't imply x^2 > y^2

SomeonecoolLovesMaths wrote: Wolog, let $a_1 < a_2 < a_3 < a_4$. If possible let, $|a_i - a_j| > \frac{1}{\sqrt{5}} \; \forall \; i,j \in \{1,2,3,4\} \; \; i \neq j$. Then, $$1 = a_1^2 + a_2^2 + a_3^2 + a_4^2 > a_1^2 + \left( \frac{1}{\sqrt{5}} + a_1 \right)^2 + \left( \frac{2}{\sqrt{5}} + a_1 \right)^2 + \left( \frac{3}{\sqrt{5}} + a_1 \right)^2$$$$\Longrightarrow 4a_1^2 - \left( \frac{12}{\sqrt{5}} \right) a_i + \frac{14}{5} < 1$$$$\Longrightarrow 4a_1^2 - \left( \frac{12}{\sqrt{5}} \right) a_i + \frac{9}{5} = 4 \left( a_i - \frac{3}{2 \sqrt{5}} \right)^2 < 0$$Impossible! Thus $\exists$ $i,j$ with $ 1 \leq i < j \leq 4$, such that $(a_i - a_j)^2 \leq \frac{1}{5}$. this is a 0 unfortunately (max 2) also megahertz13 is definitely the best sol in the thread so far

This was nice problem though I got it in last 5 seconds of contest so I wont get marks for it

alexanderhamilton124 wrote: SomeonecoolLovesMaths wrote: Wolog, let $a_1 < a_2 < a_3 < a_4$. If possible let, $|a_i - a_j| > \frac{1}{\sqrt{5}} \; \forall \; i,j \in \{1,2,3,4\} \; \; i \neq j$. Then, $$1 = a_1^2 + a_2^2 + a_3^2 + a_4^2 > a_1^2 + \left( \frac{1}{\sqrt{5}} + a_1 \right)^2 + \left( \frac{2}{\sqrt{5}} + a_1 \right)^2 + \left( \frac{3}{\sqrt{5}} + a_1 \right)^2$$$$\Longrightarrow 4a_1^2 - \left( \frac{12}{\sqrt{5}} \right) a_i + \frac{14}{5} < 1$$$$\Longrightarrow 4a_1^2 - \left( \frac{12}{\sqrt{5}} \right) a_i + \frac{9}{5} = 4 \left( a_i - \frac{3}{2 \sqrt{5}} \right)^2 < 0$$Impossible! Thus $\exists$ $i,j$ with $ 1 \leq i < j \leq 4$, such that $(a_i - a_j)^2 \leq \frac{1}{5}$. this is a 0 unfortunately (max 2) also megahertz13 is definitely the best sol in the thread so far thanks!

Let $|a_{i} - a_{j}|\geq m$. Therefore $|a_{i+1} - a_{i}|\geq m$. Consequently, $|a_{i} - a_{j}|\geq (i-j)m$. This imply that $$\sum (a_i -a_j)^2\geq 20m^2$$. Also, by the condition of the problem, we have $\sum (a_i -a_j)^2$ = $4\sum {a_i}^2-(\sum a_i)^2$≤ $4\sum {a_i}^2$=$4$. Together, we get $20m^2\leq 4\Rightarrow m^2\leq \frac {1}{5}$, as required.

alexanderhamilton124 wrote: this is a 0 unfortunately (max 2) no cuz this proves one case assuming all are positive

Sammy27 wrote: Suppose FTSOC that $(a_i - a_j)^2>\frac{1}{5}$ for all $ 1 \leq i < j \leq 4$. We are done if any two of $a_1, a_2, a_3, a_4$ are equal. WLOG, say $a_1>a_2>a_3>a_4$. Thus, we have $$a_1-a_2>\frac{1}{\sqrt{5}};\, a_2-a_3>\frac{1}{\sqrt{5}};\, a_3-a_4>\frac{1}{\sqrt{5}}.$$We systematically add to get $$a_1-a_3>\frac{2}{\sqrt{5}};\, a_2-a_4>\frac{2}{\sqrt{5}}; a_1-a_4>\frac{3}{\sqrt{5}}.$$Therefore, $$1=a_1^2 + a_2^2 + a_3^2 + a_4^2 > \sum_{i=0}^3 \left(a_4+\frac{i}{\sqrt{5}}\right)^2=4a_4^2+\frac{12a_4}{\sqrt{5}}+\frac{14}{5},$$which implies that $$0>4a_4^2+\frac{12a_4}{\sqrt{5}}+\frac{9}{5}=\left(2a_4+\frac{3}{\sqrt{5}}\right)^2\geq 0.$$This is a contradiction, and hence we are done. this is also a fakesolve

maths_enthusiast_0001 wrote: alexanderhamilton124 wrote: this is a 0 unfortunately (max 2) no cuz this proves one case assuming all are positive that can be proved trivially anyway noting $a_4 > a_1 + 3/sqrt5 > 1$ so max 1 mark

I'll just post the official solution, which was almost my approach until I realised I have only solved it for positive reals, also coz there are only 2 correct solutions in the entire thread. $\rule{40cm}{0.1pt}$ FTSOC suppose $|a_i-a_j|>\frac{1}{\sqrt{5}}$ for all $1\leq i<j \leq 4$. WLOG we take $a_i$ and $a_j$ as the maximum and minimum, then $a_i-a_j>\frac{3}{\sqrt{5}}$. Let the other two be $a_k$ and $a_\ell$, giving $a_k-a_\ell>\frac{1}{\sqrt{5}}$, we have $$1=a_i^2+a_j^2+a_k^2+a_\ell^2\geq \frac{(a_i-a_j)^2}{2}+\frac{(a_k-a_\ell)^2}{2}>\frac{1}{2}\left(\frac{9}{5}+\frac{1}{5}\right)=1$$which is a contradiction.

starchan wrote: We may order $a_1 \leq \dots \leq a_5$. There is no $a_5$, typo