373. Extension of Chebyshev's inequality.

by Virgil Nicula, May 5, 2013, 1:21 AM

Chebyshev's inequality.

$\boxed{\left|\begin{array}{c} a_1\le a_2\le\ \ldots\ \le a_{n-1}\le a_n\\\\ b_1\le b_2\le\ \ldots\ \le b_{n-1}\le b_n\end{array}\right|\ \implies\ \left(a_1+a_2+\ \ldots\ +a_n\right)\left(b_1+b_2+\ \ldots\ +b_n\right)\le n\left(a_1b_1+a_2b_2+\ \ldots\ +a_nb_n\right)}$ ,

i.e. $\frac {a_1+a_2+\ \ldots\ +a_n}{n}\cdot\frac {b_1+b_2+\ \ldots\ +b_n}{n}\le\frac {a_1b_1+a_2b_2+\ \ldots\ +a_nb_n}{n}$ .

Proof. $\sum_{j=1}^n\sum_{k=1}^n\left(a_k-a_j\right)\left(b_k-b_j\right)\ge 0\iff$ $\sum_{j=1}^n\left(\sum_{k=1}^na_kb_k-b_j\sum_{k=1}^na_k-a_j\sum_{k=1}^nb_k+na_jb_j\right)\ge 0\iff$

$2n\sum_{k=1}^na_kb_k-2\sum_{k=1}^na_k\sum_{k=1}^nb_k\ge 0\iff$ $\boxed{\sum_{k=1}^na_k\sum_{k=1}^nb_k\le n\sum_{k=1}^na_kb_k}$ .

PP. Prove that if $a_k\in \mathbb R$ for any $k\in \overline{1,n}$ , then $\prod_{k=1}^n\left(1+a_k\right)\le 2^{n-1}\cdot \left(1+\prod_{k=1}^na_k\right)$ .

Proof. Prove easily that $\boxed{(x-1)(y-1)\ge 0\iff (1+x)(1+y)\le 2(1+xy)}$ . Denote $P_s=\prod_{k=1}^sa_k\ ,\ s\in\overline{1,n}$ . Observe that $P_1=a_1$ and

$a_{s+1}P_s=P_{s+1}\ ,\ s\in\overline{1,n-1}$ . Therefore, $\left\|\begin{array}{ccc} \left(1+a_2\right)\left(1+P_1\right) & \le & 2\left(1+P_2\right)\\\\ \left(1+a_3\right)\left(1+P_2\right) & \le & 2\left(1+P_3\right)\\\\ \left(1+a_4\right)\left(1+P_3\right) & \le & 2\left(1+P_4\right)\\\\ -------- & -- & ------\\\\ \left(1+a_n\right)\left(1+P_{n-2}\right) & \le & 2\left(1+P_{n-1}\right)\\\\ \left(1+a_n\right)\left(1+P_{n-1}\right) & \le & 2\left(1+P_n\right)\end{array}\right\|\bigodot$ $\implies$ $\boxed{\prod_{k=1}^n\left(1+a_k\right)\le 2^{n-1}\cdot \left(1+\prod_{k=1}^na_k\right)}$ .

An easy extension. Let $A=\left(a_{ij}\right)$ be a real matrix with $m$ rows and $n$ columns so that $(\forall )\ i\in\overline{1,m}\implies$

$0<a_{i1}\le a_{i2}\le\ \ldots\ \le a_{ij}\le\ \ldots\ \le a_{in}$ . Then $\boxed{\prod_{i=1}^m\sum_{j=1}^na_{ij}\le n^{m-1}\sum_{i=1}^m\prod_{j=1}^n a_{ij}}$ .

Particular cases.

$1\blacktriangleright\ A=\left[\begin{array}{cc} b_1 & a_1\\\\ b_2 & a_2\\\\ -- & --\\\\ b_{m-1} & a_{m-1}\\\\ b_m & a_m\end{array}\right]$ , where $(\forall )\ k\in\overline{1,m}\ ,\ a_k\ge $ $b_k\ \implies\ \prod_{k=1}^m\left(a_k+b_k\right)\le 2^{m-1}\cdot \left(\prod_{k=1}^ma_k+\prod_{k=1}^mb_k\right)$ .

$2\blacktriangleright$ $A=\left(a_{ij}\right)$ and $(\forall )\ j\in\overline{1,n}\ ,\ (\forall )\ i\in\overline{1,m}$ we have $a_{ij}=a_j\implies$ $\left(\sum_{j=1}^na_j\right)^m\le n^{m-1}\sum_{j=1}^na_j^m$ , i.e. $\left(\frac {a_1+a_2+\ \ldots\ a_{n-1}+a_n}{n}\right)^m\le\frac {a_1^m+a_2^m+\ \ldots\ +a_{n-1}^m+a_n^m}{n}$ .

$3\blacktriangleright\ (\forall )\ k\in$ $\overline{1,n}\ ,\ x_k>0\ \implies$ $A\left(x_k\ ,\ k\in\overline{1,n}\right)=$ $\frac 1n\cdot\sum_{k=1}^nx_k\ge$ $ \frac {n}{\sum\limits_{k=1}^n\frac 1{x_k}}=$ $H\left(x_k\ ,\ k\in\overline{1,n}\right)$ .

$4\blacktriangleright$ In any acute $\triangle ABC$ exists $\sum \frac {a^2}{b^2+c^2-a^2}\ge 3$ and $\sum\tan A\ge \frac {4S}{R^2}$ . Indeed, $a^2\le b^2\le c^2\iff$ $\frac {1}{b^2+c^2-a^2}\le\frac {1}{c^2+a^2-b^2}\le\frac {1}{b^2+c^2-a^2}$ . Therefore,

$\sum\frac {a^2}{b^2+c^2-a^2}\ge$ $\frac 13\sum a^2\sum\frac {1}{b^2+c^2-a^2}=$ $\frac 13\sum \left(b^2+c^2-a^2\right)\sum\frac {1}{b^2+c^2-a^2} \ge\frac 13\cdot 9=3$ . Remark that $S=\sum[BOC]=\sum\frac {R^2\sin 2A}{2}=$ $2R^2\prod\sin A$ .

Prove easily $\sum\tan A=4S\cdot \sum\frac {1}{b^2+c^2-a^2}\ge 4S\cdot \frac 9{\sum \left(b^2+c^2-a^2\right)}=4S\cdot \frac 9{\sum a^2}\ge\frac {4S}{R^2}\implies$ $\sum\tan A\ge\frac {4S}{R^2}$ . I"ll used the inequality $a^2+b^2+c^2\le 9R^2$ .

$5\blacktriangleright\ \sqrt[2n-1]{1-x}+\sqrt[2n-1]{1+x}\le 2$ , where $|x|<1$ and $n\in\mathbb N^*$ . Indeed, denote $a=\sqrt[2n-1]{1-x}\ge 0$ and $b=\sqrt[2n-1]{1+x}\ge 0$ .Observe that

$a^{2n-1}+b^{2n-1}=(1-x)+(1+x)=2$ and $\left(\frac {a+b}{2}\right)^{2n-1}\le$ $ \frac {a^{2n-1}+b^{2n-1}}{2}=1\implies$ $a+b\le 2\implies$ $\sqrt[2n-1]{1-x}+\sqrt[2n-1]{1+x}\le 2$ .

Examples. For any positive numbers $\{a,b,c,d\}$ we have $(a+b)^4\le $ $8\left(a^4+b^4\right)\ ;\ (a+b+c)^2\le$ $3\left(a^2+b^2+c^2\right)\ ;\ (a+b+c+d)^3\le $ $16\left(a^3+b^3+c^3+d^3\right)$ .

PP1. Let $ABC$ be an acute triangle. Prove that exists the inequality $\frac {b+c}{b^2+c^2-a^2}+\frac {b^2}{c^2+a^2-b^2}+\frac {c^2}{a^2+b^2-c^2}\ge \frac 12\left(\frac 1a+\frac 1b+\frac 1c\right)$ .

Proof. Prove easily that $\tan B+\tan C\ge\frac {4}{\cot B+\cot C}=\frac {4h_a}{a}\implies$ $a\left(\tan B+\tan C\right)\ge \frac {8S}{a}\implies$ $\sum (b+c)\tan A=$ $\sum a\left(\tan B+\tan C\right)\ge$

$8S\sum\frac 1a\implies$ $\sum \frac {4S(b+c)}{b^2+c^2-a^2}\ge 8S\sum\frac 1a\implies$ $\frac {b+c}{b^2+c^2-a^2}\ge \frac 12\left(\frac 1a+\frac 1b+\frac 1c\right)$ . I"ll use the identity $\boxed{4S=\left(b^2+c^2-a^2\right)\tan A}$ .
This post has been edited 43 times. Last edited by Virgil Nicula, Nov 15, 2015, 2:07 PM

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69. Position of x1, x2 w.r.t. a given real number k.
68. Some nice applications of "pole/polar".
67. Remarkable algebraic/geometrical inequalities (1).
66. Remarkable perpendicularities in a triangle.
65. Parallel to BC through I .
64. Problems of the type "instant" (at most one minute).
63. A nice problem with radical center (livetolove212).
62. Nice and easy problem with a right triangle.
61. IMO Shortlist 2009 (very nice problem and its proof).
60. Mediteranean M.C. 2010 Problem 3.
59. IMO 2010, problem 4.
58. A cevian and two incircles.
57. Opinii si comentarii relativ la inv. rom. II
56. A extremum problem with areas.
55. Two important circles - mixtilinear incircle/excircle.
54. Cinci grade de libertate.
53. An remarkable concurrency.
52. Harmonical division.
51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.
51. A nice problem with perpendicularity from Jaime.
50. IMAC 2010 seniors, the first day.
49. The midpoint of a cevian in an acute triangle.
+ June 2010
48. An inequality in a triangle : h_a+max{b,c} ...
47. Interesting and difficult identities in a triangle.
46. Outside of a quadrilateral.
45. \cos B+\cos C=1 <==> nice property.
44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.
43. A "slicing" problem with a parameter.
42. IMAC 2010 Problema 2.
41. ONM 2010 Calarasi Problema 3.
+ May 2010
40. Lema lui Haruki.
39. A nice and remarkable problem with Brocard's point.
38. Minimum area of triangle touching with semicircle.
37. Own extension of the Steiner-Lehmus' problem.
36. ONM (juniori) - 2010 Iasi, problema 4.
35. Nice problem from Arqady (Eduard_Tur).
34. IMO Shortlist 2002 geometry problem 7.
+ April 2010
32. Linkuri diverse.
31.Some nice metrical problems.
29. Inegalitate integrala de la Kunny.
28. A fost odată ...
27.Problem of Darij Grinberg (I - centroid of triangle DEF).
26. Outside of a triangle.
25. Geometric equations.
24. A fine and cool inequalities.
23. A very nice exponential inequality (own - from CRUX).
22. Some interesting limits (sequence/function).
21 bis. Some problems with complex numbers.
21. Probleme de geometrie (Yetty & I).
20. O ineg.cu suma de AG/AI .
19. Own proposed problems in CRUX Mathematicorum - Canada.
18. O problema a lui Toshio Seimiya.
17. O inegalitate "tare" intr-un triunghi.
16. Length of the (interior) angle bisector (10 proofs).
15. Inequalities stronger than Panaitopol's.
14. Opinii si comentarii relativ la inv. rom.
13. Inegalitatea de la Sorin Borodi.
12. Inegalitatea I. Maftei & S. Radulescu (2).
11. Inegalitatea I. Maftei & S. Radulescu (1).
10. Walker's inequality and its extension.
9. Inegalitatea HO+HA+HB+HC >= 3R .
8. The first problem Shortlist ONM 2010 Romania.
7. Some interesting "slicing" problems.
6. Interesting problems from JBMO.
5. Theoretical preliminary for inequalities.
4. Remarkable geometrical inequalities (2).
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    by MrMustache, Nov 11, 2020, 4:52 PM

  • 372554 views!

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    by piphi, Jun 10, 2020, 11:44 PM

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    by CSPAL, May 27, 2020, 4:17 PM

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    by OlympusHero, May 14, 2020, 8:43 PM

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About Owner
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  • Joined: Jun 22, 2005
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  • Blog created: Apr 20, 2010
  • Total entries: 456
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