Problem 4 (second day)

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darij grinberg

6555 posts

darij grinberg

#1 h Jul 13, 2004, 2:46 PM • 11 Y

Y by Adventure10, chessgocube, megarnie, fluff_E, Mango247, bjump, buddyram, cubres, and 3 other users

Let $n \geq 3$ be an integer. Let $t_1$ , $t_2$ , ..., $t_n$ be positive real numbers such that $n^2 + 1 > \left( t_1 + t_2 + \cdots + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \cdots + \frac{1}{t_n} \right).$ Show that $t_i$ , $t_j$ , $t_k$ are side lengths of a triangle for all $i$ , $j$ , $k$ with $1 \leq i < j < k \leq n$ .

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Peter Scholze

644 posts

Peter Scholze

#2 h Jul 13, 2004, 2:54 PM • 5 Y

Y by chessgocube, Adventure10, Illuzion, Mango247, cubres

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

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pbornsztein

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pbornsztein

#3 h Jul 13, 2004, 4:12 PM • 5 Y

Y by anantmudgal09, Adventure10, chessgocube, Mango247, sabkx

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.

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jmerry

12096 posts

jmerry

#4 h Jul 13, 2004, 5:33 PM • 16 Y

Y by GammaBetaAlpha, Polynom_Efendi, ETS1331, Adventure10, oneteen11, Illuzion, Mango247, aidan0626, EquationTracker, RobertRogo, and 6 other users

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.

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Kantor

21 posts

Kantor

#5 h Jul 13, 2004, 5:40 PM • 3 Y

Y by GammaBetaAlpha, Adventure10, Mango247

it is, indeed, a very elegant solution jmerry!

Kantor

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harazi

5526 posts

harazi

#6 h Jul 14, 2004, 10:57 AM • 3 Y

Y by Adventure10, Mango247, jolynefag

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.

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Anh Cuong

431 posts

Anh Cuong

#7 h Jul 14, 2004, 4:20 PM • 2 Y

Y by Adventure10, Mango247

If my computation is not wrong then it should be

n^2+(2\sqrt10-6)n+19-6\sqrt{10}

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harazi

5526 posts

harazi

#8 h Jul 15, 2004, 11:54 AM • 3 Y

Y by Adventure10, Mango247, jolynefag

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

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Valentin Vornicu

7301 posts

Valentin Vornicu

#9 h Jul 18, 2004, 9:13 PM • 4 Y

Y by adityaguharoy, Adventure10, Mango247, jolynefag

btw, the creator of this problem is Hojoo Lee

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billzhao

827 posts

billzhao

#10 h Jul 19, 2004, 1:40 AM • 13 Y

Y by threehandsnal, Mateescu Constantin, Wizard0001, Adventure10, Periwinkle27, Jasurbek, Illuzion, Mango247, ehuseyinyigit, Bluecloud123, FairyBlade, and 2 other users

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).

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vinoth_90_2004

301 posts

vinoth_90_2004

#11 h Jul 19, 2004, 4:19 AM • 2 Y

Y by Adventure10, Mango247

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.

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harazi

5526 posts

harazi

#12 h Jul 19, 2004, 8:30 AM • 2 Y

Y by Adventure10, Mango247

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.

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Sung-yoon Kim

324 posts

Sung-yoon Kim

#13 h Jul 21, 2004, 3:52 AM • 2 Y

Y by Adventure10, Mango247

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.

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heartwork

308 posts

heartwork

#14 h Jul 21, 2004, 12:03 PM • 2 Y

Y by Adventure10, Mango247

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.

This post has been edited 2 times. Last edited by heartwork, Aug 3, 2004, 6:36 PM

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heartwork

308 posts

heartwork

#15 h Jul 21, 2004, 12:37 PM • 2 Y

Y by Adventure10, Mango247

Problem may be generalised as it follows:
Instead of a triangle, use a polygon of "given"

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

from our

real numbers can be the side lengths of a polygon with

edges. Solution is basically the same.

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shendrew7

795 posts

shendrew7

#83 h Apr 27, 2024, 2:29 AM

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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. $\blacksquare$

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RedFireTruck

4221 posts

RedFireTruck

#84 h May 7, 2024, 4:48 AM

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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.

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blueprimes

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blueprimes

#86 h Jun 22, 2024, 1:30 PM

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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.

This post has been edited 2 times. Last edited by blueprimes, Jun 24, 2024, 1:53 PM

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dudade

139 posts

dudade

#88 h Jul 31, 2024, 4:15 AM

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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.

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jolynefag

125 posts

jolynefag

#89 h Aug 5, 2024, 11:50 AM

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

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SomeonesPenguin

128 posts

SomeonesPenguin

#90 h Aug 5, 2024, 1:01 PM • 1 Y

Y by zzSpartan

storage

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numbertheory97

42 posts

numbertheory97

#91 h Aug 23, 2024, 11:50 PM

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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 $n = 3$ .

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ezpotd

1263 posts

ezpotd

#92 h Aug 26, 2024, 3:24 AM

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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.

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GeorgeRP

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GeorgeRP

#93 h Aug 27, 2024, 9:20 PM • 1 Y

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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:
$n^2> AB+t_nB+\frac{A}{t_n}\geq n^2 \text{ \Large{ \lightning }} \text{ with which we are done}$

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N3bula

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N3bula

#94 h Sep 24, 2024, 1:37 AM

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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.

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eg4334

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eg4334

#95 h Nov 29, 2024, 8:53 PM

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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.

This post has been edited 2 times. Last edited by eg4334, Nov 29, 2024, 8:56 PM

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Maximilian113

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Maximilian113

#96 h Dec 3, 2024, 11:51 PM

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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

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Marcus_Zhang

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Marcus_Zhang

#97 h Mar 23, 2025, 12:32 AM

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IMO P4

This post has been edited 1 time. Last edited by Marcus_Zhang, Mar 23, 2025, 3:46 AM

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cubres

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cubres

#98 h Apr 24, 2025, 8:46 PM

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Yapping
Storage - grinding IMO problems

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Ilikeminecraft

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Ilikeminecraft

#99 h Apr 25, 2025, 7:01 AM

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Assume there exists a sequence $t_i$ such that $t_1 \geq t_2 + t_3.$ Then, we have that:
$\begin{align*} & (t_1 + t_2 + \cdots + t_n)\left(\frac1{t_1} + \frac1{t_2} + \cdots + \frac1{t_n}\right) \\ & \geq (t_1 + t_2 + t_3)\left(\frac1{t_1} + \frac1{t_2} + \frac1{t_3}\right) + (t_1 + t_2 + t_3)\left(\frac1{t_4} + \cdots + \frac1{t_n}\right)\\ & \qquad + (t_4 + \cdots + t_n)\left(\frac1{t_1} + \frac1{t_2} + \frac1{t_3}\right) + (n - 3)^2 \\ & \geq (n - 3)^2 + 2(3n - 9) + 3 + (t_1 + t_2 + t_3)\left(\frac1{t_1} + \frac1{t_2} + \frac1{t_3}\right) \end{align*}$ Thus, all that is left is to prove that $(t_1 + t_2 + t_3)\left(\frac1{t_1} + \frac1{t_2} + \frac1{t_3}\right) \geq 10$ Expanding, we have that $3 + \frac{t_1}{t_2} + \frac{t_3}{t_2} + \frac{t_2 + t_3}{t_1} + \frac{t_1}{t_3} + \frac{t_2}{t_3}$ Now, differentiate w.r.t. $t_1.$ We get $f(t_1) = \frac1{t_2} + \frac1{t_3} - (t_2 + t_3)\frac1{t_1^2}.$ We claim that $t_1^2 \geq t_2t_3.$ However, this is clearly true because $t_1^2 \geq t_2^2 + t_3^2 + 2t_2t_3,$ while $t_2^2 + t_3^2 + t_2 t_3 \geq 0.$ Thus, $f(t_1)$ is increasing as $t_1$ increases. Thus, we can assume that $t_1 = t_2 + t_3.$ This clearly implies that $f(t_1) \geq 10.$ Hence, we are done.

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