253. An old algebraic inequality.

by Virgil Nicula, Mar 13, 2011, 6:46 AM

Proposed problem 1. Let $a,$ $b,$ $c$ be positive real numbers. Prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{(b+c-a)(c+a-b)(a+b-c)}{2abc} \le 2.$

Proof. It is equivalently with $\frac 32\ \le\ \boxed{\ \sum\frac {a}{b+c}\ \le\ 2-\frac rR\ }$ . The right inequality is equivalent to $s^2 \le 6R^2+2Rr-r^2$ which follows from Gerretsen's inequality :

$s^2 \le 4R^2+4Rr+3r^2$ since $R \ge 2r$ . Can show that $\boxed{\sum\frac a{b+c}+\frac 12+\frac rR\ \stackrel{(1)}{\le}\ \sum\frac a{b+c}+\frac{ab+bc+ca}{a^2+b^2+c^2}\ \stackrel{(2)}{\le}\ \frac 52}$ . The inequality $(1)$ is equivalently with

$\frac 12\ +\ \frac rR\ \le\ \frac{ab+bc+ca}{a^2+b^2+c^2}$ $\Longleftrightarrow$ $\frac{R+2r}{2R}\ \le\ \frac{p^2+r^2+4Rr}{2p^2-8Rr-2r^2}$ $\Longleftrightarrow$ $p^2\ \le\ 4R^2+5Rr+r^2$ . Compare with Gerretsen's inequality, i.e.

$p^2\ \le\ 4R^2+4Rr+3r^2$ ramain to show that $4R^2+4Rr+3r^2\ \le\ 4R^2+5Rr+r^2\ \Longleftrightarrow\ 2r\ \le\ R\ \Longleftrightarrow$ Euler's inequality. For the inequality $(2)$ can

write $\sum\ \frac{a}{b+c}\ =$ $\frac{\sum\ a(a+b)(a+c)}{\prod\ (b+c)}=$ $\frac{a^3+b^3+c^3+(a+b+c)(ab+bc+ca)}{(a+b+c)(ab+bc+ca)-abc}$ . Thus, $\boxed{\sum\frac{a}{b+c}=\frac{2p^2-2Rr-2r^2}{p^2+r^2+2Rr}}$ and the inequality $(2)$

becomes $\frac{2p^2-2Rr-2r^2}{p^2+r^2+2Rr}\ \le\ \frac 52\ -\ \frac{p^2+r^2+4Rr}{2p^2-8Rr-2r^2}$ $\Longleftrightarrow$ $\frac{2p^2-2Rr-2r^2}{p^2+r^2+2Rr}\le$ $\frac{2p^2-12Rr-3r^2}{p^2-r^2-4Rr}$ $\Longleftrightarrow$ $\ldots\Longleftrightarrow$

$0\le 3p^2r+2p^2R-5r^3-28Rr^2-32R^2r$ $\Longleftrightarrow\ \frac{r(32R^2+28Rr+5r^2)}{2R+3r}\ \le\ p^2$ . Compare with the Gerretsen's inequality, i.e. $16Rr-5r^2\ \le\ p^2$

remain to prove that $\frac{r(32R^2+28Rr+5r^2)}{2R+3r}\le$ $16Rr-5r^2$ $\Longleftrightarrow$ $R\ \ge\ 2r$ , what is truly.

Remarks.

$\odot$ The inequality $(2)$ appears in Secrets in inequalities (vol. I), pag. 202, where has another proof. From $(2)$ results $\boxed{\sum\frac a{b+c}+\frac rR\le 2}\ (\ast)$ .

$\odot\ \sum\frac{a^2}{b+c}=\sum\frac{b^2+c^2-2bc\cos A}{b+c}=$ $\frac{(b+c)^2-4bc\cos^2\frac A2}{b+c}=$ $4p-\sum\frac{4p(p-a)}{b+c}$ $\Longleftrightarrow$ $\sum\frac{a^2}{b+c}=4p\ -$

$2p\sum\frac{b+c-a}{b+c}$ $\Longleftrightarrow$ $\boxed{\sum\frac{a^2}{b+c}=2p\left(\sum\frac a{b+c}-1\right)}\ (\ast\ast)$ . Using the relation $(\ast)$ obtain $\boxed{\sum\frac{a^2}{b+c}\le \frac{2p(R-r)}R}$ .

$\odot$ From the inequality $(2)$ and the identity $(\ast\ast)$ obtain that $\sum\frac {a^2}{b+c}\le (a+b+c)\left(\frac 32-\frac{ab+bc+ca}{a^2+b^2+c^2}\right)\le\frac{2p(R-r)}R$ .


PP2. Prove that $\{x,y,z\}\subset\mathbb R^*_+\ \implies$ $\left(x+y+z\right)^2\left(yz+zx+xy\right)^2\le 3\left(y^2+yz+z^2\right)\left(z^2+zx+x^2\right)\left(x^2+xy+y^2\right)\ .$

Proof 1. $\left\{\begin{array}{c} 4\left(x^{2} + xy + y^{2}\right)\ge 3\left(x + y\right)^2\\\\ 4\left(y^{2} + yz + z^{2}\right)\ge 3\left(y + z\right)^2\\\\ 4\left(z^{2} + zx + x^{2}\right)\ge 3\left(z + x\right)^2\end{array}\right\|\bigodot\implies$ $64\left(y^{2} + yz + z^{2}\right)\left(z^{2} + zx + x^{2}\right)\left(x^{2} + xy + y^{2}\right)\ge $ $27\left(x + y\right)^{2}\left(y + z\right)^{2}\left(z + x\right)^{2}\ (*)\ .$ Observe that

$x\left(y - z\right)^{2} + y\left(z - x\right)^{2} + z\left(x - y\right)^{2}\ge 0\iff$ $9\left(x + y\right)\left(y + z\right)\left(z + x\right)\ge 8\left(x + y + z\right)\left(xy + yz + zx\right)\iff$ $81\left(x + y\right)^{2}\left(y + z\right)^{2}\left(z + x\right)^{2}\ge $

$64\left(x + y + z\right)^{2}\left(yz + zx + xy\right)^{2}\iff$ $\left(x + y + z\right)^{2}\left(yz + zx + xy\right)^{2}\le\frac {3}{64}\cdot 27\left(x + y\right)^{2}\left(y + z\right)^{2}\left(z + x\right)^{2}\ \stackrel{(*)}{\le}\ 3$ $\left(y^2 + yz + z^2\right)$ $\left(z^2 + zx + x^2\right)$ $\left(x^2 + xy + y^2\right)\ .$

From this chain obtain the required inequality, i.e. $\{x,y,z\}\subset\mathbb R^*_+\ \implies$ $\boxed{\left(x + y + z\right)^2 \left(yz + zx + xy\right)^2 \leq 3\left(y^2 + yz + z^2\right)\left(z^2 + zx + x^2\right)\left(x^2 + xy + y^2\right)}\ .$

Proof 2. $(\exists )\ \triangle ABC$ with $BC=a,CA=b,AB=c$ and the Fermat's point $M$ in it such that $MA=x\ ,\ MB=y\ ,\ MC=z$ , $\widehat{AMB}=\widehat{BMC}=\widehat{CMA}=120^{\circ}\ .$ Have

$\left\{\begin{array}{c} c^{2}=x^{2}+y^{2}+xy\\\\ b^{2}=z^{2}+x^{2}+zx\\\\ a^{2}=y^{2}+z^{2}+yz\end{array}\right\|$ and $\left\{\begin{array}{c} xy+yz+zx=\frac {4S}{\sqrt 3}\\\\ 2(x+y+z)^{2}=a^{2}+b^{2}+c^{2}+4S\sqrt 3\end{array}\right\|$ , where $S$ is the area of $\triangle ABC\ .$ Using the relation $4RS=abc$ we obtain that the required inequality

is equivalently with $a^{2}+b^{2}+c^{2}+4S\sqrt 3\leq 18R^{2}\ .$ But this is truly because $\left\{\begin{array}{cc} 4S\sqrt 3\le a^2+b^2+c^2 & (1)\\\\ a^2+b^2+c^2\le 9R^2 & (2)\end{array}\right\|\implies$ $a^{2}+b^{2}+c^{2}+$ $4S\sqrt 3\ \stackrel{(1)}{\leq}\ 2\left(a^2+b^2+c^2\right)\ \stackrel{(2)}{\le}\ 18R^{2}\ .$
This post has been edited 41 times. Last edited by Virgil Nicula, Nov 22, 2015, 11:05 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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