Let n be an integer greater than 1 and let $x_1, x_2, . . . , x_n$ be real numbers such that $|x_1| + |x_2| + ... + |x_n| = 1$ and $x_1 + x_2 + ... + x_n = 0$ Prove that $\left| \frac{x_1}{1}+\frac{x_2}{2}+\cdots+\frac{x_n}{n} \right| \leq \frac{1}{2} \left(1-\frac{1}{n}\right)$
Problem
Source: 2006 singapore NTST
Tags: Inequality, rearrangement inequality
achen29
30.03.2018 19:29
just to make sure; the RHS is $\frac{n}{2(n-1)}$
achen29
30.03.2018 19:50
achen29 wrote: BEAUTIFUL PROBLEM. I outline the proof. Please correct me if I am wrong!
Step 1
Prove that $\frac{n}{2(n-1)} \ge \frac{1}{2} $
Step 2
Use C-S on the LHS of the inequality to obtain
$LHS \leq \sqrt{(x_{1} ^ 2 + x_{2} ^ 2 + ...+ x_{n} ^ 2)(\frac{1}{1^2} + \frac{1}{2^2} +...+ \frac{1}{n^2})} $
Step 3
Bound the above to its maximum; the first should be bounded to smaller numbers\fractions for larger values of n; smaller values can be pure casework. The second is the well-known $\frac{\pi^2}{6}$; which you can also bound to 2.
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Click to reveal hidden textStep 4
Taking the square root; we see that what was obtained in Step 2 is indeed less than $ \frac{1}{2} $
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Click to reveal hidden textStep 5
Celebrate proving this monster, or correct any mistakes lol
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dominicleejun
02.04.2018 20:42
post #2 is wrong. equality happens when $(x_1,x_n)=(1/2,-1/2)$ or $(-1/2,1/2)$
achen29
02.04.2018 20:54
Yeah; and we know that the sum of the reciprocal of squares is less than or equal to $ \frac{ \pi ^2 }{6} $
WolfusA
07.04.2018 13:21
$LHS \leq \sqrt{(x_{1} ^ 2 + x_{2} ^ 2 + ...+ x_{n} ^ 2)(\frac{1}{1^2} + \frac{1}{2^2} +...+ \frac{1}{n^2})} \le \sqrt{\frac{1}{2}\cdot(\frac{1}{1^2} + \frac{1}{2^2} +...+ \frac{1}{n^2})}\le \sqrt{\frac{\pi^2}{12}}$
Mathskidd
16.04.2018 04:40
WolfusA wrote: $LHS \leq \sqrt{(x_{1} ^ 2 + x_{2} ^ 2 + ...+ x_{n} ^ 2)(\frac{1}{1^2} + \frac{1}{2^2} +...+ \frac{1}{n^2})} \le \sqrt{\frac{1}{2}\cdot(\frac{1}{1^2} + \frac{1}{2^2} +...+ \frac{1}{n^2})}\le \sqrt{\frac{\pi^2}{12}}$ Dear wolfusA, would you please how to prove $$x_{1} ^ 2 + x_{2} ^ 2 + ...+ x_{n} ^ 2\leq \frac 12$$thanks
achen29
16.04.2018 08:40
Mathskidd wrote: WolfusA wrote: $LHS \leq \sqrt{(x_{1} ^ 2 + x_{2} ^ 2 + ...+ x_{n} ^ 2)(\frac{1}{1^2} + \frac{1}{2^2} +...+ \frac{1}{n^2})} \le \sqrt{\frac{1}{2}\cdot(\frac{1}{1^2} + \frac{1}{2^2} +...+ \frac{1}{n^2})}\le \sqrt{\frac{\pi^2}{12}}$ Dear wolfusA, would you please how to prove $$x_{1} ^ 2 + x_{2} ^ 2 + ...+ x_{n} ^ 2\leq \frac 12$$thanks Cauchy-Schwarz!
Mathskidd
16.04.2018 08:53
achen29 wrote: Mathskidd wrote: WolfusA wrote: $LHS \leq \sqrt{(x_{1} ^ 2 + x_{2} ^ 2 + ...+ x_{n} ^ 2)(\frac{1}{1^2} + \frac{1}{2^2} +...+ \frac{1}{n^2})} \le \sqrt{\frac{1}{2}\cdot(\frac{1}{1^2} + \frac{1}{2^2} +...+ \frac{1}{n^2})}\le \sqrt{\frac{\pi^2}{12}}$ Dear wolfusA, would you please how to prove $$x_{1} ^ 2 + x_{2} ^ 2 + ...+ x_{n} ^ 2\leq \frac 12$$thanks Cauchy-Schwarz! are you sure?
WolfusA
16.04.2018 12:20
Mathskidd wrote: Dear wolfusA, would you please how to prove $$x_{1} ^ 2 + x_{2} ^ 2 + ...+ x_{n} ^ 2\leq \frac 12$$thanks $|x_1| + |x_2| + ... + |x_n| = 1\implies \sum_{l=1}^{n}x_l^2+2\sum_{1\le k< j\le n} |x_kx_j|=1$ $x_1 + x_2 + ... + x_n = 0\implies \sum_{l=1}^{n}x_l^2+2\sum_{1\le k< j\le n} x_kx_j=0$ Adding acquired equalities $1\ge 2\sum_{l=1}^{n}x_l^2+2\sum_{1\le k< j\le n} (x_kx_j+ |x_kx_j|)\ge 2\sum_{l=1}^{n}x_l^2$ because for all $t\in R$ we have $|t|\ge -t$
WolfusA
16.04.2018 12:22
achen29 wrote: Mathskidd wrote: Dear wolfusA, would you please how to prove $$x_{1} ^ 2 + x_{2} ^ 2 + ...+ x_{n} ^ 2\leq \frac 12$$thanks Cauchy-Schwarz! C-S was the first step. Proof of $x_{1} ^ 2 + x_{2} ^ 2 + ...+ x_{n} ^ 2\leq \frac 12$ has nothing to do with C-S.
Mathskidd
16.04.2018 12:29
WolfusA wrote: Mathskidd wrote: Dear wolfusA, would you please how to prove $$x_{1} ^ 2 + x_{2} ^ 2 + ...+ x_{n} ^ 2\leq \frac 12$$thanks $|x_1| + |x_2| + ... + |x_n| = 1\implies \sum_{l=1}^{n}x_l^2+2\sum_{1\le k< j\le n} |x_kx_j|=1$ $x_1 + x_2 + ... + x_n = 0\implies \sum_{l=1}^{n}x_l^2+2\sum_{1\le k< j\le n} x_kx_j=0$ Adding acquired equalities $1\ge 2\sum_{l=1}^{n}x_l^2+2\sum_{1\le k< j\le n} (x_kx_j+ |x_kx_j|)\ge 2\sum_{l=1}^{n}x_l^2$ because for all $t\in R$ we have $|t|\ge -t$ thanks a lot.