solution by induction: the hardest part is to show that it works for n=3. so assume $t_1\geq t_2\geq t_3$. it is easy to see that the right is strictly increasing in $t_3$. so if we show that the reverse inequality holds for $t_1=t_2+t_3$, we are done. a quick calculation is enough to show that(it is too boring to post). but if we know is true for n-1, then let's do the following computation: $((t_1+...+t_{n-1})+t_n)(\frac{1}{t_1}+...+\frac{1}{t_{n-1}}+\frac{1}{t_n})\geq (t_1+...+t_{n-1})(\frac{1}{t_1}+...+\frac{1}{t_{n-1}})+2n-1$, by AM-GM and AM-HM. so we get the inequality reduces to that one for n-1 variables. Peter

It reminds me the following problem from China 1987/88 : Let $n \geq 3$ be an integer. Suppose the inequality $(a_1^2 + a_2^2 + ... + a_n^2)^2 > (n-1)(a_1^4 + ... a_n^4)$ holds for some positive real numbers $a_1,...a_n$. Prove that $a_i,a_j,a_k$ are the three sides of some triangle for all $i<j<k$. Even there is an elegant direct solution, the most common way is induction. Pierre.

When I saw this, my immediate thought was- an elegant direct solution. The inequality $n^2 + 1 > \left( t_1 + t_2 + ... + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + ... + \frac{1}{t_n} \right)$ reduces to $\displaystyle{n^2+1>\sum_{j=1}^n \sum_{k=1}^n{\frac{t_j}{t_k}}}$ $\displaystyle{n^2+1>n+\sum_{1 \le j<k \le n}{\frac{t_j}{t_k}+\frac{t_k}{t_j}}}$ Each of the terms in the last sum is at least 2, so if the sum of some $m$ terms is at least $2m+1$, the whole sum will be at least $n^2-n+1$ and the inequality will be violated. Suppose $t_i+t_j \le t_k$. Then $\frac{t_i}{t_k}+\frac{t_j}{t_k} \le 1$ If we can show that $a+b+\frac{1}{a}+\frac{1}{b} \ge 5$ when $a+b \le 1$, we will be done by the argument above. First, we have $a+\frac{1}{a}=\frac{a^2+1}{a} \ge \frac{(a-\frac{1}{2})^2+a+\frac{3}{4}}{a} \ge 1+\frac{3}{4a}$ and $b+\frac{1}{b} \ge 1+\frac{3}{4b}$ so $a+b+\frac{1}{a}+\frac{1}{b} \ge 2+\frac{3}{4}(\frac{1}{a}+\frac{1}{b})$ By AM-HM, $\frac{2}{\frac{1}{a}+\frac{1}{b}} \le \frac{a+b}{2} \le \frac{1}{2}$, so $\frac{1}{a}+\frac{1}{b} \ge 4$ and $a+b+\frac{1}{a}+\frac{1}{b} \ge 5$ In fact the case n=3 is sufficient to prove the desired fact for all n - we can simply pair off the other terms and only worry about the ones in our non-triangle.

it is, indeed, a very elegant solution jmerry! Kantor

Well, I just adore this problem. So easy, beautiful and elegant. But try to think about the following problem, which I think it's much more difficult and interesting: For iven n>2, find the greatest constant k (which may depend on n) such that if positive reals $ x_1,...,x_n$ verify $ k>(x_1+...+x_n)(\frac{1}{x_1}+...+\frac{1}{x_n}) $ then any three of them are sides of a triangle. I think I've managed to solve this problem and if my solution is correct, then it's a really hard and nice problem.

If my computation is not wrong then it should be n^2+(2\sqrt10-6)n+19-6\sqrt{10}

Yes, it is $ (n+\sqrt{10}-3)^2 $. I think this one is much more interesting.

btw, the creator of this problem is Hojoo Lee

Another solution: Note that \[ (t_1+ \cdots + t_n) ( \frac{1}{t_1} + \cdots + \frac{1}{t_n}) = n^2 + \sum_{i<j} (\sqrt{\frac{t_i}{t_j}} - \sqrt{\frac{t_j}{t_i}} )^2 \] And so if three of the terms satisfy $a \geq b + c$ then \[{{ (\sqrt{a/b} - \sqrt{b/a})^2 + (\sqrt{a/c} - \sqrt{c/a})^2 = \frac{ (a-b)^2}{ab} } + \frac{ (a-c)^2}{ac} } \geq \frac{(a-b)c}{ab} + \frac{(a-c)b}{ac} = c/b + b/c -(b+c)/a \geq 2 - 1 = 1 \] And the result follows immediately. (During the IMO, I accidentally forgot the square root, costing me a huge chunk of marks).

just wanted to say that this is *really* nice ... btw, anh cuong/harazi could you please post the solution to the harder version (where n^2+1 is replaced by ( n + :sqrt: 10 - 3 ) ^ 2 )? thanks.

Of course, Hojoo Lee is the official proposer. But I knew the case n=3 some four months ago, when I tried to create a problem for the junior TST in Romania.

It's my solution in the 2004 IMO. By Cauchy-Schwartz inequality, (t_1+t_2+...+t_n)(1/t_1+1/t_2+...+1/t_n)=(3\times(t_1+t_2+t_3)/3 +t_4+...+t_n)(3\times(1/t_1+1/t_2+1/t_3)/3 +1/t_4+...+1/t_n)\geq(3\times \sqrt((t_1+t_2+t_3)/3\times(1/t_1+1/t_2+1/t_3)/3) +1+...+1)^2 So n^2+1>(\sqrt((t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3)) +n-3)^2 => (t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3)<(\sqrt(n^2+1) -(n-3))^2 As n\geq3, it follows (\sqrt(n^2+1) -(n-3))^2<10 <=> 2n^2 -6n<(2n-6) \sqrt(n^2+1) So (t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3)<10 Let t_1 \geq t_2 \geq t_3. Then t_1+t_2 \geq t_3, t_1+t_3 \geq t_2 Assume that t_1>t_2+t_3. Then (t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3)-((t_2+t_3)+t_2+t_3)(1/(t_2+t_3) +1/t_2+1/t_3) =(t_1-t_2-t_3)(1/t_2+1/t_3-1/t_1) \geq 0 so by Cauchy-Schwartz inequality, 10>(t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3) \geq 2(t_2+t_3)(1/(t_2+t_3)+1/t_2+1/t_3) <=> 4>(t_2+t_3)(1/t_2+1/t_3) \geq 4 and contradiction.

harazi wrote: ...find the greatest constant k (which may depend on n) such that if positive reals x_1,x_2,...,x_n verify k(n)>(x_1+x_2+...+x_n)(1/x_1+1/x_2+...+1/x_n) then any three of them are sides of a triangle. harazi wrote: Yes, it is (n+\sqrt{10}-3)^2 . I think this one is much more interesting. Consider f(x_1,x_2,...,x_n)=(x_1+x_2+...+x_n)(1/x_1+1/x_2+...+1/x_n) where x_1<=x_2<=...<=x_n. Greatest constant k(n)=(n+\sqrt{10}-3)^2 might be found elementary. From the solution of Sung-yoon Kim: Sung-yoon Kim wrote: ...By Cauchy-Schwartz inequality, .......... So n^2+1>(\sqrt((x_1+x_2+x_3)(1/x_1+1/x_2+1/x_3))+(n-3)^2 => (x_1+x_2+x_3)(1/x_1+1/x_2+1/x_3)<(\sqrt(n^2+1) -(n-3))^2 instead of (n^2+1) , use that k(n) and instead of x_3 , use x_n: (x_1+x_2+x_n)(1/x_1+1/x_2+1/x_n)<(\sqrt(k(n)) -(n-3))^2 Assume x_1+x_2<=x_n then: (x_1+x_2+x_n)(1/x_1+1/x_2+1/x_n)>=10 , so we must have 10>=(\sqrt(k(n)) -(n-3))^2, which leads to the wanted answer. Indeed if: (x_1+x_2+x_n)(1/x_1+1/x_2+1/x_n)<(\sqrt(k(n)) -(n-3))^2, then: x_1+x_2>x_n , so for any 1<= i, j, k <= n : x_i+x_j>x_k . Of course, there is a quite simple (but not elementary) "forged" solution, using optimization theory: k(n)=min f(x_1,x_2,...,x_n) , when x_1<=x_2<=...<=x_n and x_1+x_2<=x_n , we may get: k(n)=f(1,1,2\sqrt(2/5),...,2\sqrt(2/5),2), where 2\sqrt(2/5) appears (n-3) times. I think it's pretty interesting.

Problem may be generalised as it follows: Instead of a triangle, use a polygon of "given" M edges , M>=3 so if the same inequality holds for any n>=M positive reals (of course with something like k(n,M) in the RHS) then any M from our n real numbers can be the side lengths of a polygon with M edges. Solution is basically the same.

The solution I came up with is fairly distinct from what I see here, so I thought I'd share it. I assume t_1 >= t_2 + t_3 and try to show (sum t_i) (sum 1/t_i) >= n^2 + 1. First increase t_3 and decrease t_1, fixing t_1+t_3, until t_1 = t_2+t_3. This fixes (sum t_i) and decreases (sum 1/t_i). Thus, it suffices to consider the case where t_1 = t_2 + t_3. Now, let X = sum t_i (i >= 4), Y = sum t_j (j >= 4): (sum t_i)(sum 1/t_i) = 0.5(t_1 + t_2 + t_3 + X)(2/t_1 + Y) + 0.5(t_1 + t_2 + t_3 + X)(2/t_2 + 2/t_3 + Y) = 0.5(t_1 + t_1 + X)(1/t_1 + 1/t_1 + Y) + 0.5(t_2 + t_2 + t_3 + t_3 + X)(1/t_2 + 1/t_2 + 1/t_3 + 1/t_3 + Y) >= 0.5(n-1)^2 + 0.5(n+1)^2 by Cauchy-Schwarz = n^2 + 1 The smoothing trick I used at the start can be used further to get the optimal solution. We can assume t_1 = t_2 + t_3, then we can assume t_2 = t_3, and then then we can assume t_4 = t_5 = t_6 = ... = t_n. Now, we just have 2 variables. Multiply it out and use AM-GM.

Here is a related problem: Find the smallest f(n) such that if (t1 + ... + tn)(1/t1 + ... + 1/tn) < f(n) then for every i, j we have ti/tj <2

My solution was with induction like the others, but a little different: First I proved it for n=3, then I said suppose it's true for n and not for n+1 (take $t_1 \leq ... \leq t_n$) so I worked out for $t_{n+1} \geq t_1 + t_2$ and after applying many times that $\frac ab + \frac ba \geq 2$ and that $\frac ca + \frac cb = \frac c{a+b} . (2+\frac ab + \frac ba) \geq 4.\frac c{a+b}$ combined with our $t_{n+1} \geq t_1 + t_2$, I got that when we really minimalize the RHS we get: $n^2 + 2n + 2 > n^2 + 2n + 2$ so contradiction. Same for the case $t_{n+1} + t_1 \leq t_2$.

I thought, why wouldn't I post my solution, it is more or less elegant... Let's first do the case $n = 3$. Proof by contradiction: if, for example, $c \geq a + b$ then we have \[(a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \geq (a + b + c)\left(\frac{4}{a + b} + \frac{1}{c}\right) = 5 + \frac{4c}{a + b} + \frac{a + b}{c} \geq 5 + 5 \sqrt[5]{\frac{c^4}{(a + b)^4} \cdot \frac{a + b}{c}} \geq 5 + 5 = 10\] which proves the case $n = 3$. In the general case we use again a proof by contradiction - this <a href="http://www.ntsearch.com/search.php?q=time&v=56">time</a> we assume WLOG that if the conclusion is false, then $t_1 + t_2\leq t_3$. Let's take $t_1 = a$, $t_2 = b$, $t_3 = c$, $\sum_{i = 4}^nt_i = S$ and $\sum_{i = 4}^n \frac{1}{t_i} = T$. Then the given product equals \[(a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) + S\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) + T(a + b + c) + ST.\]By Cauchy we have $ST \geq (n - 3)^2$ and we also have \[ S\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) + T(a + b + c) \geq 2 \sqrt{ST(a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)} \geq 6(n - 3) \] so by the case $n = 3$ the product is at least \[ (n - 3)^2 + 6(n - 3) + 10 = n^2 + 1 \] which is the desired contradiction. (You see, no induction... )

quite elegant indeed.

Suppose not, then we have $a,b,c\in t_i$ such that $a+b\le c$. But note that the terms in the RHS besides \[S=\frac ac+\frac bc+\frac ca+\frac cb\]will add up to at least $n^2-4$ by AM-GM, so we have $S<5$. WLOG $c=1$, so \[S=\left(1+\frac{1}{ab}\right)(a+b).\]For a fixed $s=a+b$, $1+\frac{1}{ab}$ is minimized when $a=b$. Therefore, \[S\ge 2a+\frac{2}{a}\ge 2\cdot\frac12+\frac{2}{\frac12}=5,\]contradiction. $\blacksquare$

10 mohs alg is harder than 20 mohs geo Suppose for the sake of contradiction there exists $a,b,c \in t_i$, such that $c= a+ b+ \epsilon$ for $\epsilon \ge 0$ then by cauchy $$\left(t_1 + t_2 + \dots + t_n\right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} \right) \ge \left(\sqrt{(a+b+a+b+\epsilon)\left(\frac{1}a+ \frac{1}b+ \frac{1}{a+b+\epsilon} \right)} + (n-3) \right)^2$$We claim $$(a+b+a+b+\epsilon)\left(\frac{1}a+ \frac{1}b+ \frac{1}{a+b+\epsilon} \right) > 10$$ This is true because it equals $$3+\frac{a}{b}+\frac{b}{a}+\frac{\epsilon}{a}+\frac{\epsilon+a+b}{a}+ \frac{\epsilon +a+b}{b}+ \frac{a+b}{a+b+\epsilon}$$$$\ge 9+ \frac{\epsilon}{a}+ \frac{\epsilon}{b}+1 -\frac{\epsilon}{a+b+\epsilon}$$$$\ge 10+ \epsilon(\tfrac{1}{a}+\tfrac{1}{b}- \tfrac{1}{a+b+\epsilon})$$$$>10$$Now $$\left(t_1 + t_2 + \dots + t_n\right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} \right) > (\sqrt{10}+n-3)^2$$Which is greater than $n^2+1$ for $n\ge 3$ contradiction $\square$

Assume FTSOC that $t_n \geq t_1 + t_2$. We claim that $\frac{t_1}{t_n} + \frac{t_1}{t_2} + \frac{t_n}{t_1} + \frac{t_n}{t_2} + \frac{t_2}{t_n} \geq 7$ Notice that as $t_n$ increases the expression decreases. Thus, we just need to prove that when $t_n = t_1+t_2$ we have the equality case. This is easy since we have \[\frac{t_1}{t_n} + \frac{t_1}{t_2} + \frac{t_n}{t_1} + \frac{t_n}{t_2} + \frac{t_2}{t_n} = 3 + \frac{2t_1}{t_2} + \frac{2t_2}{t_1} \geq 7\]Now we can use AM-GM on all of the other terms in the expansion of our product to see the result. $\blacksquare$. Very similar to @asdf334

We first prove the case $n=3$. Assume FTSOC we have $(t_1,t_2,t_3) = (a,b,a+b+k)$, where $k \ge 0$. Then \begin{align*} 3^2+1 &> (t_1+t_2+t_3)\left(\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3}\right) \\ &\ge 5 + \frac{t_3}{t_1} + \frac{t_3}{t_2} + \frac{t_1}{t_2} + \frac{t_1}{t_3} \\ &= 7 + \frac{c+k}{b} + \frac{b+k}{c} + \frac{b+c}{b+c+k} \\ &\ge 10 + k\left(\frac 1b + \frac 1c - \frac{1}{b+c+k}\right) \\ &\ge 10, \end{align*} contradiction. We now show the case $n \ge 4$. Assume again FTSOC $t_1, t_2, t_3$ do not form the sides of a triangle. Then \begin{align*} n^2+1 &> \left(t_1 + t_2 + \ldots + t_n\right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_n} \right) \\ &= 10 + (n-3)^2 + \left(t_1+\ldots+t_3\right)\left(\frac{1}{t_4}+\ldots+\frac{1}{t_n}\right) + \left(t_4+\ldots+t_n\right) \left(\frac{1}{t_1}+\ldots+\frac{1}{t_3}\right) \\ &\ge 10 + (n-3)^2 + 2\sqrt{10(n-3)^2} \\ &= \left(n+\sqrt{10}-3\right)^2 \\ &\ge n^2+1, \end{align*} contradiction. $\blacksquare$

Assume WLOG that $t_1\le t_2\le \dots\le t_n$. Claim: $a\ge b+c$ for $a,b,c>0$ means that $\frac{b+c}{a}+\frac{a}{b}+\frac{a}{c}\ge 5$. Proof: Let $a=b+c+d$. Then, we wish to prove that $(1-\frac{d}{b+c+d})+2+\frac{b}{c}+\frac{c}{b}+\frac{d}{b}+\frac{d}{c}\ge 5$. By AM-GM, $\frac{b}{c}+\frac{c}{b}\ge 2$ so it suffices to prove that $\frac{d}{b}+\frac{d}{c}\ge \frac{d}{b+c+d}$, which is true because $\frac1b+\frac1c\ge \frac1{b+c}$. Notice that for $n=3$, we just need to prove that $t_3\ge t_1+t_2$ implies that $(t_1+t_2+t_3)(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3})\ge 10$. Expanding gives $$(t_1+t_2+t_3)(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3})=3+\frac{t_1}{t_2}+\frac{t_2}{t_1}+\frac{t_1+t_2}{t_3}+\frac{t_3}{t_1}+\frac{t_3}{t_2}\ge 3+2+5=10$$as desired. Assume that for some $n-1\ge 3$, the assertion is true. We will now prove that the assertion is true for $n$. If there exists $1\le i<j<k\le n-1$ such that $t_i$, $t_j$, $t_k$ don't form the side lengths of a triangle, then $$(t_1+t_2+\dots+t_{n-1})(\frac{1}{t_1}+\frac{1}{t_2}+\dots+\frac{1}{t_{n-1}})\ge (n-1)^2+1$$and $$t_n(\frac1{t_1}+\frac1{t_2}+\dots+\frac1{t_{n-1}})+(t_1+t_2+\dots+t_{n-1})\frac{1}{t_n}+1\ge 2n-1$$by AM-GM so $$(t_1+t_2+\dots+t_{n})(\frac{1}{t_1}+\frac{1}{t_2}+\dots+\frac{1}{t_{n}})\ge n^2+1$$as desired. Otherwise, assume that $t_n\ge t_1+t_2$. By AM-GM and our claim, $$(t_1+t_2+\dots+t_{n})(\frac{1}{t_1}+\frac{1}{t_2}+\dots+\frac{1}{t_{n}})\ge n+2\binom{n}{2}-4+\frac{t_1+t_2}{t_n}+\frac{t_n}{t_1}+\frac{t_n}{t_2}\ge n^2+1$$as desired. We are done by induction.

For the sake of contradiction, assume that three of the $t_i$ are equal to $a$, $b$, $a + b + x$ where $a$, $b$ are positive reals and $x$ is nonnegative. By Cauchy-Schwarz, we have $$(t_1 + t_2 + \dots + t_n) \left( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} \right) \ge \left(n - 3 + \sqrt{(2a + 2b + x) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{a + b + x} \right)} \right)^2.$$Now $$\sqrt{(2a + 2b + x) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{a + b + x} \right)} = \sqrt{6 + \frac{2a}{b} + \frac{2b}{a} + x \left( \frac{1}{a} + \frac{1}{b} - \frac{1}{a + b + x} \right) } \ge \sqrt{10}$$by AM-GM. To reach contradiction it is now sufficient to show $(n - 3 + \sqrt{10})^2 \ge n^2 + 1 \iff n \ge 3$ which is true. We are done.

FTSOC, suppose $t_1 > t_2 + t_3$. Therefore, we want to show \begin{align*} n^2 + 1 &> \sum_{1 \leq i<j \leq n} \left(\dfrac{t_i}{t_j} + \dfrac{t_j}{t_i}\right) + n \\ &= \left(\left(\dfrac{t_1}{t_2}+\dfrac{t_2}{t_1}\right) + \left(\dfrac{t_2}{t_3}+\dfrac{t_3}{t_2}\right) + \left(\dfrac{t_3}{t_1}+\dfrac{t_1}{t_3}\right)\right) + \sum_{4\leq i<j \leq n} \left(\dfrac{t_i}{t_j} + \dfrac{t_j}{t_i}\right)+n \\ &\geq \left(\left(\dfrac{t_1}{t_2}+\dfrac{t_2}{t_1}\right) + \left(\dfrac{t_2}{t_3}+\dfrac{t_3}{t_2}\right) + \left(\dfrac{t_3}{t_1}+\dfrac{t_1}{t_3}\right)\right) + \left(n^2 - n - 6\right) + n. \end{align*}Thus, we want \begin{align*} 7 > \left(\dfrac{t_1}{t_2}+\dfrac{t_2}{t_1}\right) + \left(\dfrac{t_2}{t_3}+\dfrac{t_3}{t_2}\right) + \left(\dfrac{t_3}{t_1}+\dfrac{t_1}{t_3}\right) \end{align*}if $t_1 > t_2 + t_3$. By AM-GM-HM, note that \begin{align*} \left(\dfrac{t_1}{t_2}+\dfrac{t_2}{t_1}\right) + \left(\dfrac{t_2}{t_3}+\dfrac{t_3}{t_2}\right) + \left(\dfrac{t_3}{t_1}+\dfrac{t_1}{t_3}\right) &= \left(\dfrac{t_2}{t_1} + \dfrac{t_3}{t_1}\right) + \left(\dfrac{t_1}{t_2} + \dfrac{t_1}{t_3}\right) + 2 \\ &\geq \left(\dfrac{t_2}{t_1} + \dfrac{t_3}{t_1}\right)+ \dfrac{4}{\left(\dfrac{t_1}{t_2} + \dfrac{t_1}{t_3}\right)} + 2 = k + \tfrac{4}{k} + 2. \end{align*}Since $k < 1$, then we must have that $k + \tfrac{4}{k} + 2 \leq 7$ which is clearly a contradiction. Thus, for all $i$, $j$, and $k$, there exists a triangle with side lengths $t_i$, $t_j$, and $t_k$, as desired.

Elegant problem which is still being solved by people these days. So beautiful!

Really nice problem! I think I accidentally ended up proving the stronger version. Solution. Without loss of generality, suppose to the contrary that $t_1 \geq t_2 + t_3$. Then \[t_1 + t_2 + t_3 = \frac25 t_1 + \frac35 t_1 + (t_2 + t_3) \geq \frac25 t_1 + \frac85(t_2 + t_3)\]and \[\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} \geq \frac{1}{t_1} + \frac{4}{t_2 + t_3}\]by Cauchy-Schwarz, so we have \[(t_1 + t_2 + \dots + t_n)\left(\frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n}\right)\]\[ \geq \left(\frac25 t_1 + \frac85(t_2 + t_3) + t_4 + \dots + t_n\right)\left(\frac{1}{t_1} + \frac{4}{t_2 + t_3} + \frac{1}{t_4} + \dots + \frac{1}{t_n}\right)\]\[ \overset{\text{C-S}}{\geq} \left(\sqrt{\frac25} + \sqrt{\frac{32}{5}} + (n - 3)\right)^2 = \left(n + \sqrt{10} - 3\right)^2.\]But $f(x) = \left(x + \sqrt{10} - 3\right)^2 - x^2$ is strictly increasing, so $f(x) \geq f(3) = 1$, a contradiction. $\square$ Remark. (Motivation) Why $2/5$ and $3/5$? I originally tried using $1/2$'s, which was very close but broke slightly for $n = 3$. But if we use arbitrary coefficients $x$ and $1 - x$, it suffices to maximize the function \[f(x) = \sqrt x + 2\sqrt{2 - x},\]which by C-S or similar methods is greatest when $x = 2/5$. This fits perfectly when $n = 3$.

Sort $t_n$. Assume to the contrary that there exists some tuple $(t_1 \cdots t_n)$ such that the inequality is satisfied and $t_1 + t_2 \le t_n$. We show that if $n \ge 4$ we can form a similar tuple with $n -1$ elements. Of course, just remove $t_3$, the decrease on the left is $2n - 1$, the decrease on the right is $t_3 \frac{1}{t_3} + \sum_{i \neq 3} \frac{t_i}{t_3} + \frac{t_3}{t_i} \ge 2n - 1$ by AM-GM, so the inequality is preserved. Now we can just reduce to the $n = 3$ case, where we MUST have $10 > 3 + \sum_{sym} \frac{t_i}{t_j}$. Since $\frac{t_3}{t_1} + \frac{t_1}{t_3}, \frac{t_3}{t_2} + \frac{t_2}{t_3}$ are increasing for $t_3 > t_1, t_2$, if the inequality holds true for $t_3 > t_1 + t_2$, it must hold for $t_3 = t_1 + t_2$. Now plugging this value of $t_3$ gives $7 > 2(\frac{t_1}{t_2} + \frac{t_2}{t_1}) + 3$, which is NEVER true by AM-GM, so the original inequality can never hold in the conditions described.

We will solve the problem by induction on n. Base case: n=3. We have that: $$10>(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \Leftrightarrow 7abc>\sum{a^2b}$$WLOG we can assume $a$ is the largest. Thus we need to show $a<b+c$. If we rewrite as a quadratic function in respect to $a$ (considering $b$ and $c$ as parameters) we have: $$0>a^2(b+c)+a(b^2+c^2-7bc)+(b^2c+c^2b)=f(a)$$We however know: $$ b+c>a \Leftrightarrow \begin{cases} f(b+c)\geq0 \\ b+c>\frac{7bc-b^2-c^2}{2(b+c)} \end{cases} \Leftrightarrow \begin{cases} 2(b+c)^3-8(b+c)bc\geq0\\ 3b^2-3bc+3c^2>0 \end{cases} \text{ which are both true, finishing the base}$$ IH: for n-1 IS: for n. Let $A=t_1+t_2+\cdots+t_{n-1}, B=\frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_{n-1}}$.WLOG we can assume $t_1\leq t_2 \leq \cdots \leq t_n$. We know that $n^2+1>(A+t_n)(B+\frac{1}{t_n}) \Leftrightarrow n^2-Bt_n-\frac{A}{t_n}>AB$. We will first show that $Bt_n+\frac{A}{t_n}\geq2n-2$. This follows from the fact: $$Bt_n+\frac{A}{t_n}=\sum_{1\leq i \leq n-1}{\frac{t_n}{t_i}+\frac{t_i}{t_n}} \overset{AM-GM}{\geq} 2n-2$$From this it directly follows that $(n-1)^2+1>AB$, implying that $t_i, t_j, t_k$ are the sides of a triangle for $1\leq i<j<k\leq n-1$. We are now left to show that $t_n<t_1+t_2$ from which the desired will follow (because $t_1, t_2$ are the smallest). For simplicity let $t_1=x, t_2=y$. FTSOC let $t_n=x+y+k$ for some $k\geq0$. We know that by Cauchy-Schwarz $AB\geq(n-1)^2$. We will now show that $t_nB+\frac{A}{t_n}\geq 2n-1$. $$t_nB+\frac{A}{t_n}= \frac{t_n}{t_1}+\frac{t_1}{t_n}+\frac{t_2}{t_n}+\frac{t_n}{t_2}+(\frac{t_3}{t_n}+\cdots+\frac{t_{n-1}}{t_n})+(\frac{t_n}{t_3}+\cdots+\frac{t_n}{t_{n-1}}) \overset{\text{AM-GM}}{\geq} $$$$ \geq \frac{a+b+k}{a}+\frac{a}{a+b+k}+\frac{a+b+k}{b}+\frac{b}{a+b+k}+2n-6 \geq \frac{a+b}{a}+\frac{a}{a+b}+\frac{a+b}{b}+\frac{b}{a+b}+2n-6=$$$$=\frac{(a+b)^2}{ab}+2n-5\overset{\text{AM-GM}}{\geq}2n-1$$From here it follows that: $$n^2> AB+t_nB+\frac{A}{t_n}\geq n^2 \text{ \Large{ \lightning }} \text{ with which we are done}$$

Expanding gives us: \[\left(t_1 + t_2 + \dots + t_n\right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} \right) = n+\sum_{i, j \in \{1, 2, \dots, n\}, i\neq j} \frac{t_i}{t_j} +\frac{t_j}{t_i}\]Thus we get if for any $i$, $j$ and $k$ that $\sum_{sym}\frac{t_i}{t_j}\geq 7$ we get a contradiction, thus it suffices to prove that if we have $a+b\leq c$ that $\sum_{sym}\frac{a}{b}\geq 7$, thus I will prove that that sum is minimised when $a+b=c$, we get that proving that is equivalent to proving \[\frac{1}{a}+\frac{1}{b}-\frac{a+b}{c(c+k)}\geq 0\]When $a+b=c$ and $k$ is a positive real. Thus the above is equivalent to: \[\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\geq 0\]\[\frac{b+a}{ab}-\frac{1}{a+b}\]The above is minimised when $a=b$, thus its equivalent to: \[\frac{2a}{a^2}-\frac{1}{2a}\geq 0\]Which is clearly true.

WLOG $t_1 \leq t_2 \leq \dots \leq t_n$, and FTSOC let $t_i + t_j < t_k$. The condition then gives by Cauchy that $$n^2 +1 \geq \left( n -3 + \sqrt{3+\sum \frac{t_i}{t_j} } \right)^2$$where the summation is taken over all ordered pairs in $t_i, t_j, t_k$. Now notice that $n^2 +1 \geq (n - 3 + \sqrt{10})^2$ has no integer solutions for $n \geq 3$, so we just need to prove that $$\frac{t_i}{t_j} + \frac{t_i}{t_k} + \frac{t_j}{t_i} + \frac{t_j}{t_k} + \frac{t_k}{t_i} + \frac{t_k}{t_j} \geq 7 $$$$\frac{t_i}{t_k} + \frac{t_j}{t_k} + \frac{t_k}{t_i} + \frac{t_k}{t_j} \geq5$$by AMGM. Also, $(\frac{1}{t_i} + \frac{1}{t_j})(t_i + t_j) \geq 4$ by Cauchy so this reduces into $$\frac{t_i+t_j}{t_k} + \frac{4t_k}{t_i+t_i} \geq 5$$which is obvious.

We show by induction on $n$ that if there exists at least one triple $(t_i, t_j, t_k)$ that are not the sides of a triangle, then $$n^2+1 \leq (t_1+t_2+\cdots+t_n)\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_n} \right).$$$\text{Base Case: }$ For $n=3,$ we must show that if $t_1, t_2, t_3$ cannot be the sides of a triangle, then $$10 \leq (t_1+t_2+t_3)\left( \frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3} \right).$$WLOG, assume that $t_1 \leq t_2 \leq t_3,$ then we have that $t_3 > t_1+t_2.$ Also, because our inequality is homogenous, assume that $t_3=1.$ Letting $x=t_1+t_2,$ it suffices to show that $$7 \leq \frac{t_1}{t_2}+\frac{t_2}{t_1}+x+\frac{1}{t_1}+\frac{1}{t_2}.$$But by AM-GM, $\frac{t_1}{t_2}+\frac{t_2}{t_1} \geq 2,$ so it suffices to show that $$x+\frac{1}{t_1}+\frac{1}{t_2} \geq 5.$$But, $x \leq t_3=1 \implies (x-1)(x-4) \geq 0 \implies 5 \leq x+\frac{4}{x} \leq x+\frac{1}{t_1t_2} \leq x+\frac{1}{t_1}+\frac{1}{t_2}$ by $2$ applications of AM-GM at the end. Therefore, our base case is proven. $\text{Inductive Step: }$ Now, assume that for $n=k-1 \geq 3$ our proposition holds. Consider the sequence $t_1, t_2, \cdots t_k$ of positive real numbers. Then if there is at least one triple, say WLOG $(t_1, t_2, t_3),$ which are not the sides of a triangle then we have by our inductive hypothesis that \begin{align*} (t_1+t_2+\cdots+t_k)\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_k} \right) &=(t_1+t_2+\cdots+t_{k-1})\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_{k-1}} \right) + \sum_{i=1}^{k-1} \left( \frac{t_i}{t_k}+\frac{t_k}{t_i} \right)+1 \\ &\geq \left((k-1)^2+1\right)+(k-1)\cdot 2+1 \text{ by AM-GM} \\ &= k^2+1. \end{align*}Therefore, our induction is complete. QED Now, for the sake of a contradiction assume that there exists $t_i, t_j, t_k$ which are not the sides of a triangle, and $$n^2+1 > (t_1+t_2+\cdots+t_n)\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_n} \right).$$But by our induction above, we have that $$n^2+1 \leq (t_1+t_2+\cdots+t_n)\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_n} \right),$$a contradiction. Therefore, all $t_i, t_j, t_k$ are the sides of a triangle. QED