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2023 IMO Problems/Problem 4: Difference between revisions

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==Solution==
==Solution==
We first solve for <math>a_1.</math> <math>a_1 = \sqrt{(x_1)(\frac{1}{x_1})} = \sqrt{1} = 1.</math>
We first solve for <math>a_1.</math> <math>a_1 = \sqrt{(x_1)(\frac{1}{x_1})} = \sqrt{1} = 1.</math>
Now we solve for <math>a_{n+1}</math> in terms of <math>a_n</math> and <math>x.</math> <math>a_{n+1}^2\\ = (\sum^{n+1}_{k=1}x_k)(\sum^{n+1}_{k=1}\frac1{x_k}) \\= (x_{n+1}+\sum^{n}_{k=1}x_k)(\frac{1}{x_{n+1}}+\sum^{n}_{k=1}\frac1{x_k}) \\= 1 + \frac{1}{x_{n+1}} \sum^{n}_{k=1}x_k+x_{n+1}\sum^{n}_{k=1}\frac1{x_k} + (\sum^{n}_{k=1}x_k)(\sum^{n}_{k=1}\frac1{x_k})\\=1+\frac{1}{x_{n+1}} \sum^{n}_{k=1}x_k+x_{n+1}\sum^{n}_{k=1}\frac1{x_k}+a_n^2</math>
Now we solve for  
By AM-GM, <math>a_{n+1}^2 \ge 1+a_n^2 + 2 \sqrt{(\frac{1}{x_{n+1}} \sum^{n}_{k=1}x_k)(x_{n+1} \sum^{n}_{k=1} \frac1{x_k})} = 1 + a_n^2 + 2a_n = (a_n+1)^2 \\ </math>
<math>a_{n+1}</math> in terms of <math>a_n</math> and <math>x.</math> <math>a_{n+1}^2 \\  
= (\sum^{n+1}_{k=1}x_k)(\sum^{n+1}_{k=1}\frac1{x_k}) \\
= (x_{n+1}+\sum^{n}_{k=1}x_k)(\frac{1}{x_{n+1}}+\sum^{n}_{k=1}\frac1{x_k}) \\  
= 1+\frac{1}{x_{n+1}}\sum^{n}_{k=1}x_k+x_{n+1}\sum^{n}_{k=1}\frac1{x_k} + (\sum^{n}_{k=1}x_k)(\sum^{n}_{k=1}\frac1{x_k}) \\  
= 1+\frac{1}{x_{n+1}} \sum^{n}_{k=1}x_k+x_{n+1}\sum^{n}_{k=1}\frac1{x_k}+a_n^2</math>
 
By AM-GM, <math>a_{n+1}^2 \ge 1+a_n^2 + 2 \sqrt{(\frac{1}{x_{n+1}} \sum^{n}_{k=1}x_k)(x_{n+1} \sum^{n}_{k=1} \frac1{x_k})} = 1 + a_n^2 + 2a_n = (a_n+1)^2 \\ \text{}</math>
 
Hence, <math>a_{n+1} \ge a_n.</math> However, this is not of much use, as we can only determine that <math>a_{2023} \ge 2023. \\ \text{}</math>
 
Now, we solve for <math>a_{n+2}</math> in terms of <math>a_n</math> and <math>x.</math>
<math>a_{n+2}^2 \\
= (\sum^{n+2}_{k=1}x_k)(\sum^{n+2}_{k=1}\frac1{x_k}) \\
= (x_{n+1}+x_{n+2}+\sum^{n}_{k=1}x_k)(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}}+\sum^{n}_{k=1}\frac1{x_k}) \\
= \frac{x_{n+1}}{x_{n+1}} + \frac{x_{n+1}}{x_{n+2}} + \frac{x_{n+2}}{x_{n+1}} + \frac{x_{n+2}}{x_{n+2}} +
\frac{1}{x_{n+1}}\sum^{n}_{k=1}x_k + x_{n+1}\sum^{n}_{k=1}\frac1{x_k} + \frac{1}{x_{n+2}}\sum^{n}_{k=1}x_k + x_{n+2}\sum^{n}_{k=1}\frac1{x_k} + (\sum^{n}_{k=1}x_k)(\sum^{n}_{k=1}\frac1{x_k}) \\
= \frac{x_{n+1}}{x_{n+1}} + \frac{x_{n+1}}{x_{n+2}} + \frac{x_{n+2}}{x_{n+1}} + \frac{x_{n+2}}{x_{n+2}} +
\frac{1}{x_{n+1}}\sum^{n}_{k=1}x_k + x_{n+1}\sum^{n}_{k=1}\frac1{x_k} + \frac{1}{x_{n+2}}\sum^{n}_{k=1}x_k + x_{n+2}\sum^{n}_{k=1}\frac1{x_k} + a_n^2 \\ \text{}</math>
 
Again, by AM-GM, the above equation becomes
<math>a_{n+2}^2 \ge 4 \sqrt[4]{(\frac{x_{n+1}}{x_{n+1}})(\frac{x_{n+1}}{x_{n+2}})(\frac{x_{n+2}}{x_{n+1}})(\frac{x_{n+2}}{x_{n+2}})} +
4\sqrt[4]{
(\frac{1}{x_{n+1}}\sum^{n}_{k=1}x_k)(x_{n+1}\sum^{n}_{k=1}\frac1{x_k})(\frac{1}{x_{n+2}}\sum^{n}_{k=1}x_k)(x_{n+2}\sum^{n}_{k=1}\frac1{x_k})
}
+ a_n^2  = a_n^2+4a_n+4 = (a_n+2)^2 \\ \text{}</math>
 
Hence, <math>a_{n+2} \ge a_{n} + 2,</math> but equality is achieved only when <math>\frac{x_{n+1}}{x_{n+1}},\frac{x_{n+1}}{x_{n+2}},\frac{x_{n+2}}{x_{n+1}}, </math> and <math>\frac{x_{n+2}}{x_{n+2}}</math> are equal. They can never be equal because there are no two equal <math>x_k.</math>So <math>a_{2023} \ge a_1 + 3\times \frac{2023-1}{2} = 1 + 3033 = 3034</math>

Revision as of 00:19, 14 July 2023

Problem

Let $x_1, x_2, \cdots , x_{2023}$ be pairwise different positive real numbers such that \[a_n = \sqrt{(x_1+x_2+···+x_n)(\frac1{x_1} + \frac1{x_2} +···+\frac1{x_n})}\] is an integer for every $n = 1,2,\cdots,2023$. Prove that $a_{2023} \ge 3034$.

Solution

We first solve for $a_1.$ $a_1 = \sqrt{(x_1)(\frac{1}{x_1})} = \sqrt{1} = 1.$ Now we solve for $a_{n+1}$ in terms of $a_n$ and $x.$ $a_{n+1}^2 \\ = (\sum^{n+1}_{k=1}x_k)(\sum^{n+1}_{k=1}\frac1{x_k}) \\ = (x_{n+1}+\sum^{n}_{k=1}x_k)(\frac{1}{x_{n+1}}+\sum^{n}_{k=1}\frac1{x_k}) \\ = 1+\frac{1}{x_{n+1}}\sum^{n}_{k=1}x_k+x_{n+1}\sum^{n}_{k=1}\frac1{x_k} + (\sum^{n}_{k=1}x_k)(\sum^{n}_{k=1}\frac1{x_k}) \\ = 1+\frac{1}{x_{n+1}} \sum^{n}_{k=1}x_k+x_{n+1}\sum^{n}_{k=1}\frac1{x_k}+a_n^2$

By AM-GM, $a_{n+1}^2 \ge 1+a_n^2 + 2 \sqrt{(\frac{1}{x_{n+1}} \sum^{n}_{k=1}x_k)(x_{n+1} \sum^{n}_{k=1} \frac1{x_k})} = 1 + a_n^2 + 2a_n = (a_n+1)^2 \\ \text{}$

Hence, $a_{n+1} \ge a_n.$ However, this is not of much use, as we can only determine that $a_{2023} \ge 2023. \\ \text{}$

Now, we solve for $a_{n+2}$ in terms of $a_n$ and $x.$ $a_{n+2}^2 \\ = (\sum^{n+2}_{k=1}x_k)(\sum^{n+2}_{k=1}\frac1{x_k}) \\ = (x_{n+1}+x_{n+2}+\sum^{n}_{k=1}x_k)(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}}+\sum^{n}_{k=1}\frac1{x_k}) \\ = \frac{x_{n+1}}{x_{n+1}} + \frac{x_{n+1}}{x_{n+2}} + \frac{x_{n+2}}{x_{n+1}} + \frac{x_{n+2}}{x_{n+2}} + \frac{1}{x_{n+1}}\sum^{n}_{k=1}x_k + x_{n+1}\sum^{n}_{k=1}\frac1{x_k} + \frac{1}{x_{n+2}}\sum^{n}_{k=1}x_k + x_{n+2}\sum^{n}_{k=1}\frac1{x_k} + (\sum^{n}_{k=1}x_k)(\sum^{n}_{k=1}\frac1{x_k}) \\ = \frac{x_{n+1}}{x_{n+1}} + \frac{x_{n+1}}{x_{n+2}} + \frac{x_{n+2}}{x_{n+1}} + \frac{x_{n+2}}{x_{n+2}} + \frac{1}{x_{n+1}}\sum^{n}_{k=1}x_k + x_{n+1}\sum^{n}_{k=1}\frac1{x_k} + \frac{1}{x_{n+2}}\sum^{n}_{k=1}x_k + x_{n+2}\sum^{n}_{k=1}\frac1{x_k} + a_n^2 \\ \text{}$

Again, by AM-GM, the above equation becomes $a_{n+2}^2 \ge 4 \sqrt[4]{(\frac{x_{n+1}}{x_{n+1}})(\frac{x_{n+1}}{x_{n+2}})(\frac{x_{n+2}}{x_{n+1}})(\frac{x_{n+2}}{x_{n+2}})} + 4\sqrt[4]{ (\frac{1}{x_{n+1}}\sum^{n}_{k=1}x_k)(x_{n+1}\sum^{n}_{k=1}\frac1{x_k})(\frac{1}{x_{n+2}}\sum^{n}_{k=1}x_k)(x_{n+2}\sum^{n}_{k=1}\frac1{x_k}) } + a_n^2 = a_n^2+4a_n+4 = (a_n+2)^2 \\ \text{}$

Hence, $a_{n+2} \ge a_{n} + 2,$ but equality is achieved only when $\frac{x_{n+1}}{x_{n+1}},\frac{x_{n+1}}{x_{n+2}},\frac{x_{n+2}}{x_{n+1}},$ and $\frac{x_{n+2}}{x_{n+2}}$ are equal. They can never be equal because there are no two equal $x_k.$So $a_{2023} \ge a_1 + 3\times \frac{2023-1}{2} = 1 + 3033 = 3034$

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