135. Three strong geometrical inequalities.

by Virgil Nicula, Oct 2, 2010, 12:19 PM

PP1 (own). Let $ABC$ be a triangle for which $a\ge b$ , $a\ge c$ . Prove that $\frac {b^2+c^2}{bc}+\frac {b+c}{a}\ge \frac {1}{\sin \frac A2}\cdot\sqrt {1+\left(\frac {b-c}{2a}\right)^2}$ .

Proof. Observe that $a\ge b\ \wedge\ a\ge c$ $\Longrightarrow$ $ 2a\ge b+c\ \wedge\ A\ge 60^{\circ}$ , i.e. $1-2\cos A\ge 0$ . Thus, $2a(b+c)\ge (b+c)^2\ge 4bc$ $\Longleftrightarrow$

$2a(b+c)+bc\ge \frac {b^2c^2}{(s-b)(s-c)}+bc-\frac {b^2c^2}{(s-b)(s-c)}+4bc$ $\Longleftrightarrow$ $2a(b+c)+bc\ge \frac {b^2c^2}{(s-b)(s-c)}-\frac {bcs(s-a)}{(s-b)(s-c)}+4bc$ because

$s(s-a)+(s-b)(s-c)=bc$ . Therefore, $2a(b+c)(b-c)^2+bc(b-c)^2\ge $ $\frac {b^2c^2\left[a^2-4(s-b)(s-c)\right]}{(s-b)(s-c)}-\frac {bcs(s-a)((b-c)^2}{(s-b)(s-c)}+4bc(b-c)^2$

because $(b-c)^2=a^2-4(s-b)(s-c)$ . Thus, $2a(b+c)^2(b-c)^2+bc(b-c)^2\ge$ $ \frac {a^2b^2c^2}{(s-b)(s-c)}-4b^2c^2-\frac {bcs(s-a)(b-c)^2}{(s-b)(s-c)}+4bc(b-c)^2$

$\Longleftrightarrow$ $2a(b+c)(b-c)^2+bc(b+c)^2\ge $ $\frac {a^2b^2c^2}{(s-b)(s-c)}+bc(b-c)^2\left(4-\cot^2\frac A2\right)$ $\Longleftrightarrow$ $\frac {2(b+c)(b-c)^2}{abc}+$ $\left(\frac {b+c}{a}\right)^2\ge $

$\frac {bc}{(s-b)(s-c)}+\left(\frac {b-c}{a}\right)^2\cdot\frac {4\sin^2\frac A2-\cos^2\frac A2}{\sin^2\frac A2}$ $\Longleftrightarrow$ $\left[\frac {(b-c)^2}{bc}+\frac {b+c}{a}\right]^2-\left[\frac {(b-c)^2}{bc}\right]^2\ge \frac {1}{\sin^2\frac A2}\cdot\left[1+\left(\frac {b-c}{a}\right)^2\cdot\frac {3-5\cos A}{2}\right]$ $\Longleftrightarrow$

$\boxed{\frac {(b-c)^2}{bc}+\frac {b+c}{a}\ge\frac {1}{\sin\frac A2}\cdot\sqrt {1+\left(\frac {b-c}{a}\right)^2\cdot \frac {3-5\cos A}{2}}}\ (*)$ . From $\cos A\le\frac 12$ obtain $3-5\cos A\ge \frac 12$ .

Therefore, $\frac {b^2+c^2}{bc}+\frac {b+c}{a}\ge$ $ 2+\frac {1}{\sin \frac A2}\cdot\sqrt {1+\left(\frac {b-c}{2a}\right)^2}\ \ (1)$ . Have equality iff $b=c$ .

Remark. If $A=90^{\circ}$ , then the inequality $(1)$ becomes $\frac {a^2}{bc}+\frac {b+c}{a}\ge$ $ 2+\sqrt 2\cdot\sqrt {1+\left(\frac {b-c}{2a}\right)^2}\ge 2+\sqrt 2\ \ (2)$ . But $2+2\cdot\left(\frac {b-c}{2a}\right)^2\ge $ $\frac {9a^2-2bc}{4a^2}$ $\Longleftrightarrow$

$8a^2+2\left(a^2-2bc\right)^2\ge $ $9a^2-2bc$ $\Longleftrightarrow$ $a^2\ge 2bc$ $\Longleftrightarrow$ $(b-c)^2\ge 0$ , what is truly. Thus, the inequality $(2)$ becomes $A=90^{\circ}\implies$

$\frac {a^2}{bc}+\frac {b+c}{a}\ge$ $ 2+\frac {1}{2a}\cdot\sqrt {8a^2+(b-c)^2}\ge 2+\sqrt 2$ . In conclusion, $\boxed{\ A=90^{\circ}\ \implies\ \frac {a^2}{bc}+\frac {b+c}{a}\ \ge\ 2+\sqrt 2\ }$ (Crux Mathematicorum).


PP2. Prove that in any triangle $ABC$ the following inequality holds : $\boxed{\ \cos\frac A2+\cos\frac B2+\cos\frac C2\ \ge\ \sqrt{2\cdot\left(\sin\frac A2+\sin\frac B2+\sin\frac C2\right)^3}\ }$ .

Proof. If $X$ is an acute angle then $0<X<\frac {\pi}2\iff 0<\frac {\pi-A}2<\frac{\pi}2\iff 0<A<\pi$ . This means that $\forall\ A\in (0,\pi)$ it exists only one

$X\in\left(0,\frac {\pi}2\right)$ so that $A=\pi-2X$ . Consequently, we can make the following substitutions : $\left\|\ \begin{array}{cccc} A=\pi-2X \\ B=\pi-2Y \\ C=\pi-2Z\end{array}\ \right\|$ where $X+Y+Z=\pi$ and

$X,Y,Z\in \left(0,\frac {\pi}2\right)$ and thus the proposed inequality will be equivalent to : $\left[\ \sum_{\text{cyc}}\ \cos\left(\frac {\pi}2-X\right)\ \right]^2\ \ge\ 2\cdot\left[\ \sum_{\text{cyc}}\ \sin\left(\frac {\pi}2-X\right)\ \right]^3\ \iff$

$\left(\sum_{\text{cyc}}\ \sin X\right)^2\ \ge\ 2\cdot\left(\sum_{\text{cyc}}\ \cos X\right)^3$ . Using the well-known identities for a triangle $XYZ\ :\ \sum\ \sin X=\frac sR$ and $\sum\ \cos X=1+\frac rR$

the last inequality becomes : $s^2\ \ge\ \frac {2(R+r)^3}R\ (\ast)$ . Comparing this inequality to Walker's i.e. $\boxed{\ s^2\ \ge\ 2R^2+8Rr+3r^2\ }$ it is enough to

prove that : $2R^2+8Rr+3r^2\ \ge\ \frac {2(R+r)^3}R\iff (R-2r)(2R+r)\ge 0$ , what is obviously . Therefore, the inequality $(\ast)$ is truly.


PP3. Prove that in any $\triangle ABC$ there is the inequality $\boxed{\ \cos A\cdot\cos B+\cos B\cdot\cos C+\cos C\cdot\cos A\ \ge\ 6\cdot\cos A\cdot\cos B\cdot\cos C\ }$ .

Proof. Using $\left\|\ \begin{array}{cccc} \sum\ \cos B\cdot\cos C=\frac {s^2+r^2-4R^2}{4R^2} \\ \\ \prod\ \cos A=\frac {s^2-(2R+r)^2}{4R^2}\ \end{array}\right\|$ the given inequality becomes $\frac {s^2+r^2-4R^2}{4R^2}\ge 6\cdot\frac {s^2-(2R+r)^2}{4R^2}\iff $

$s^2\ \le\ 4R^2+\frac {24}5\cdot Rr+\frac 75\cdot r^2$ . Using Gerrensen's inequality $\boxed{\ s^2\ \le\ 4R^2+4Rr+3r^2\ }\ (1)$ and Euler's inequality $\boxed{\ R\ge 2r}\ (2)$

obtain that $s^2\ \stackrel{(1)}{\le}\ 4R^2+4Rr+3r^2=$ $\left(4R^2+\frac {24}{5}\cdot Rr+\frac 75\cdot r^2\right)-\frac 45\cdot r(R-2r)\ \stackrel{(2)}{\implies}\ s^2$ $\le 4R^2+\frac {24}{5}\cdot Rr+\frac 75\cdot r^2$ .

Remark. If $\triangle ABC$ is acute, then our inequality is equivalently with $\sum\frac {1}{\cos A}\ge 6$ , what is truly because $\sum\frac {1}{\cos A}\ge \frac {9}{\sum \cos A}=\frac {9}{1+\frac rR}=$

$\frac {9R}{R+r}\ge 6$ because $\frac {9R}{R+r}\ge 6\iff R\ge 2r$ . Otherwise, the function $f\ :\ \left(\ 0\ ,\ \frac {\pi}{2}\ \right)\ \rightarrow\ (\ 0\ ,\ \infty\ )$ , where $f(x)=\frac {1}{\cos x}$ is convex.
This post has been edited 25 times. Last edited by Virgil Nicula, Nov 23, 2015, 6:30 AM

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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!

    by mrmath0720, Sep 28, 2020, 1:11 AM

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    -piphi

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

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    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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  • Blog created: Apr 20, 2010
  • Total entries: 456
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