24. A fine and cool inequalities.

by Virgil Nicula, Apr 22, 2010, 10:10 PM

Prove that in any triangle $ABC$ there are the following inequalities :

$1.\blacktriangleright \ \ \frac {\sqrt 3}{R}\le \frac 1a+\frac 1b+\frac 1c\le \frac {\sqrt 3}{2r}\le\frac {1}{b+c-a}+\frac {1}{c+a-b}+\frac {1}{a+b-c}$ .

$2.\blacktriangleright \ \ a+b+c\le \frac {a^2+b^2}{a+b}+\frac {b^2+c^2}{b+c}+\frac {c^2+a^2}{c+a}\le 3\cdot\frac {a^2+b^2+c^2}{a+b+c}$ .

$3.\blacktriangleright \ \ \frac 1{4r^2}\ge\frac {1}{a^2}+\frac {1}{b^2}+\frac {1}{c^2}\ge \frac {1}{ab}+\frac {1}{bc}+\frac {1}{ab}\ge \frac {1}{R^2}$ (Goldstone's inequality).

Proof.

$\blacktriangleright\ a^2\ge 4(s-b)(s-c)$ a.s.o. $\implies$ $\sum\frac 1{a^2}\le \sum \frac 1{4(s-b)(s-c)}=$ $\sum\frac {s-a}{4(s-a)(s-b)(s-c)}=$ $\sum\frac {s-a}{4sr^2}=$ $\frac {\cancel s}{4\cancel sr^2}=$ $\frac 1{4r^2}\implies$ $\boxed{\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}\le\frac 1{4r^2}}\ .$

$\blacktriangleright\ \sum b^2c^2=\sum (bc)^2\ge (bc)\cdot (ca)+(ca)\cdot (ab)+(ab)\cdot (bc)=abc(a+b+c)\implies \sum b^2c^2\ge abc\cdot\sum a\ \stackrel{:\ (abc)^2}{\implies}\ \sum \frac 1{a^2}\ge \sum \frac 1{bc}\implies$ $\boxed{\frac 1{bc}+\frac 1{ca}+\frac 1{ab}\le \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}}\ .$

$\blacktriangleright\ \sum \frac 1{bc}=\frac {2s}{abc}=\frac {2s}{4Rsr}=\frac 1{2Rr}\ge \frac 1{R^2}\implies$ $\boxed{\frac 1{R^2}\le\frac 1{bc}+\frac 1{ca}+\frac 1{ab}}\ .$ In conclusion, $\boxed{\frac 1{R^2}\le \frac 1{bc}+\frac 1{ca}+\frac 1{ab}\le \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}\le\frac 1{4r^2}}\ .$ Otherwise. Direct proof of $\boxed{\frac 1{R^2}\le \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}}\ :$

$\blacktriangleright\ \left\{\begin{array}{c} 4(s-a)(s-b)\le c^2\\\\ 4(s-a)(s-c)\le b^2\end{array}\right|\bigodot\implies$ $16(s-a)sr^2\le b^2c^2\ \stackrel{:\ (abc=4Rrs)}{\iff}\ \frac {16(s-a)sr^2}{4Rsr}\le \frac {b^2c^2}{abc}\iff$ $\frac {4r(s-a)}{R}\le\frac {bc}{a}\ \stackrel{\sum}{\implies}\ \frac {4Rrs}{R^2}\le \sum \frac {bc}{a}\ \stackrel{:\ (abc=4Rrs)}{\implies}\ \boxed{\frac 1{R^2}\le \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}}\ .$


$4.\blacktriangleright \ \ 1+\frac {b+c-a}{a}+$ $\frac {c+a-b}{b}+\frac {a+b-c}{c}\le\frac {2r}{R}+$ $\frac {a}{b+c-a}+\frac {b}{c+a-b}+\frac {c}{a+b-c}$ .

$5.\blacktriangleright \ \ a^2+b^2+c^2+16r^2\le 2(ab+bc+ca)$ (easy !).

$6.\blacktriangleright$ IF $x$ , $y$ , $z$ are positive numbers, THEN $\boxed{\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge\frac{x+y+z}{\sqrt[3]{xyz}}}>3\ .$

Proof. $3\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)=$ $\left(\frac{2x}{y}+\frac{y}{z}\right)+\left(\frac{2y}{z}+\frac{z}{x}\right)+\left(\frac{2z}{x}+\frac{x}{y}\right)\ge$ $\frac{3x}{\sqrt[3]{xyz}}+\frac{3y}{\sqrt[3]{xyz}}+\frac{3z}{\sqrt[3]{xyz}}$ .

$7.\blacktriangleright$ Prove that $(\forall )$ an acute $\triangle ABC$ there is the inequality $\sum\frac 1{a^2}+\sum\frac 1{HA^2}\le \frac 1{4\rho^2}\ ,$ where $\rho$ is the length of the inradius for the orthic $\triangle DEF$ w.r.t. $\triangle ABC\ .$

Proof. Denote the orthic triangle $DEF\ ,$ where $D\in (BC)\ ,$ $E\in (CA)$ and $F\in (AB)\ .$ Prove easily that the orthocenter $H\in AD\cap BE\cap CF$ of $\triangle ABC$ is the incenter of $\triangle DEF$ and $\rho$

is the distance $\delta_{EF}(H)$ from $H$ to $EF\ .$ Thus, $EF\cdot \delta_{EF}(H)=2\cdot [EHF]=$ $HE\cdot HF\cdot\sin\widehat{EHF}\ \stackrel{(2)}{\iff}\ R\sin 2A\cdot\delta_{EF}(H)=$ $AH\cos \widehat{AHE}\cdot AH\cos\widehat{AHF}\cdot\sin\widehat{EHF}\ \stackrel{(1)}{\iff}$

$\cancel{2R}\cancel{\sin A}\cancel{\cos A}\cdot\rho =$ $\cancel{2R}\cancel{\cos A}\cos C\cdot 2R\cos A\cos B\cdot\cancel{\sin A}\iff$ $\boxed{\rho =2R\cos A\cos B\cos C}\ (*)\ .$ I"ll use the well-known relations $:\ \left\{\begin{array}{cc} \frac {BC}{\sin A}=\frac {CA}{\sin B}=\frac {AB}{\sin C}=2R & (0)\\\\ \frac {HA}{\cos A}=\frac {HB}{\cos B}=\frac {HC}{\cos C}=2R & (1)\\\\ \frac {EF}{\sin 2A}=\frac {FD}{\sin 2B}=\frac {DE}{\sin 2C}=R & (2)\end{array}\right\|$ and

$BC^2+AH^2=CA^2+BH^2=AB^2+CH^2=4R^2\ (3)\ .$ Therefore, $\sum\frac 1{a^2}+\sum\frac 1{HA^2}\le \frac 1{4\rho^2}\ \stackrel{(3)}{\iff}\ \sum\frac {4R^2}{a^2\cdot HA^2}\le \frac 1{4\rho^2}\ \stackrel{(*\wedge 0\wedge 1)}{\iff}\ \sum \frac {\cancel{4R^2}}{\cancel{4R^2}\sin^2A\cdot \cancel{4R^2}\cancel{\cos^2A}}\le $

$\frac 1{4\cdot \cancel{4R^2}\cancel{\cos^2A}\cos^2B\cos^2C}$ $\iff$ $\sum\frac {\cos^2B\cos^2C}{\sin^2A}\le \frac 14\ .$ With the substitutions $A\to \frac {\pi - A}2$ a.s.o. this inequality becomes $\frac {\cos^2\left(90^{\circ}-\frac B2\right)\cos^2\left(90^{\circ}-\frac C2\right)}{\sin^2\left(90^{\circ}-\frac A2\right)}\le\frac 14\iff$ $\frac {\sin^2\frac B2\sin^2\frac C2}{\cos^2\frac A2}\iff$

$\sum \frac {\cancel{(s-a)}(s-c)}{a\cancel c}\cdot\frac {(s-a)(s-b)}{a\cancel b}\cdot\frac {\cancel{bc}}{s\cancel{(s-a)}}\le \frac 14\iff$ $\sum\frac {(s-a)(s-b)(s-c)}{sa^2}\le\frac 14\iff$ $\sum\frac{\cancel sr^2}{\cancel sa^2}\le \frac 14\iff$ $\sum\frac 1{a^2}\le\frac 1{4r^2}\ ,$ what is true. Hence and required inequality is true.

Application (particular case). Apply the inequality 1.7 to $\triangle I_aI_bI_c$ w.r.t. $\triangle ABC\ :\ \boxed{\sum\frac 1{I_bI_c^2}+\sum\frac 1{II_a^2}\le\frac 1{4r^2}}\ .$
This post has been edited 110 times. Last edited by Virgil Nicula, Aug 22, 2017, 10:09 AM

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These are nice problems! I may try some of them and post my solutions on my blog tomorrow. Be sure to check it out! :D

by Kingofmath101, Aug 10, 2011, 2:54 AM

Own problems or extensions/generalizations of some problems which was posted here.

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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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  • Joined: Jun 22, 2005
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  • Total entries: 456
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