4. Remarkable geometrical inequalities (2).

by Virgil Nicula, Apr 19, 2010, 2:22 PM

III. Remarkable geometrical inequalities.

$\bigodot$ Let $x_k$ , $y_k$ , $k\in\overline {1,n}\ .$ Then exists the Cauchy-Bunyakovsky-Schwarz's inequality (C.B.S) $\boxed {\left(\sum_{k=1}^nx_ky_k\right)^2\le\sum_{k=1}^nx_k^2\cdot\sum_{k=1}^ny_k^2}\ .$ See here. We have equality if and only if

$\frac {x_1}{y_1}=\frac {x_2}{y_2}=\ldots =\frac {x_n}{y_n}$ with the convention $x_k=0\Longleftrightarrow y_k=0$ (see here ). Another usual forms $:\ \sum _{k=1}^n\frac {a_k^2}{b_k}\cdot\sum_{k=1}^n b_k\ge\left(\sum_{k=1}^na_k\right)^2\ ;\ \left(\sum_{k=1}^na_k\right)\left(\sum_{k=1}^nx_k\right)\ge \left(\sum_{k=1}^n\sqrt {a_kx_k}\right)^2$ .

Method 1. I"ll use the Lagrange's identity $\sum_{k=1}^nx_k^2\cdot\sum_{k=1}^ny_k^2=\left(\sum_{k=1}^nx_ky_k\right)^2+\sum_{1\le k<j\le n}\left(x_ky_j-x_jy_k\right)^2\ge\left(\sum_{k=1}^nx_ky_k\right)^2\ .$

Method 2. Consider the polynomial $f\equiv\sum_{k=1}^n\left(a_kX-b_k\right)^2=X^2\cdot \sum_{k=1}^nx_k^2-2X\cdot\sum_{k=1}^nx_ky_k+\sum_{k=1}^ny_k^2\ .$ Observe that

$(\forall )\ x\in\mathbb R\ ,\ f(x)\ge 0\ .$ Hence $\Delta'=\left(\sum_{k=1}^nx_ky_k\right)^2-\sum_{k=1}^nx_k^2\cdot\sum_{k=1}^ny_k^2\le 0$ , i.e. $\left(\sum_{k=1}^nx_ky_k\right)^2\ \le\ \sum_{k=1}^nx_k^2\cdot x\sum_{k=1}^ny_k^2\ .$

Remark. For $x_k>0$ , $k\in\overline {1,n}$ can write the C.B.S's inequality thus :

$\boxed {\ \sum_{k=1}^nx_k\cdot\sum_{k=1}^n\frac {y_k^2}{x_k}\ \ge\ \sum_{k=1}^ny_k^2\ }$ OR $\boxed {\ \sum_{k=1}^nx_ky_k\cdot\sum_{k=1}^n\frac {y_k}{x_k}\ \ge\ \left(\sum_{k=1}^ny_k\right)^2\ }$ OR $\boxed {\ \sum_{k=1}^nx_k\cdot\sum_{k=1}^nx_ky_k^2\ \ge\ \left(\sum_{k=1}^n x_ky_k\right)^2\ }$ . Indeed :

$\odot\ \ \sum_{k=1}^ny_k^2\ =$ $\sum_{k=1}^n\left(\sqrt {x_k}\cdot\frac {y_k}{\sqrt {x_k}}\right)^2\ \le$ $\sum_{k=1}^n\left(\sqrt {x_k}\right)^2\cdot\sum_{k=1}^n\left(\frac {y_k}{\sqrt {x_k}}\right)^2\ =$ $\sum_{k=1}^nx_k\cdot\sum_{k=1}^n\frac {y^2_k}{x_k}\ .$

$\odot\ \ \sum_{k=1}^nx_ky_k\cdot\sum_{k=1}^n\frac {y_k}{x_k}\ =$ $\sum_{k=1}^n\left(\sqrt {x_ky_k}\right)^2\cdot\sum_{k=1}^n\left(\sqrt{\frac {y_k}{x_k}}\right)^2\ \ge$ $\left(\sum_{k=1}^n\sqrt {x_ky_k\cdot\frac {y_k}{x_k}}\right)^2\ =$ $\left(\sum_{k=1}^ny_k\right)^2\ .$

$\odot\ \ \sum_{k=1}^nx_k\cdot\sum_{k=1}^nx_ky_k^2\ =$ $\sum_{k=1}^n\left(\sqrt {x_k}\right)^2\cdot\sum_{k=1}^n\left(y_k\sqrt{x_k}\right)^2\ \ge\ \left(\sum_{k=1}^n\sqrt {x_k}\cdot y_k\sqrt {x_k}\right)^2\ =\ \left(\sum_{k=1}^n x_ky_k\right)^2\ .$

Example. If $a$ , $b$ , $c$ are positively, then $\boxed {\ \frac {a}{b+c}+\frac {b}{c+a}+\frac {c}{a+b}\ \ge \frac 32\ }$ (the Nesbitt's imequality).

Metoda 1. $\sum\frac {a}{b+c}=\sum\frac {a^2}{a(b+c)}\ge\frac {(a+b+c)^2}{2(ab+bc+ca)}\ge \frac 32$ deoarece $(a+b+c)^2\ge 3(ab+bc+ca)\ .$

Metoda 2. Se foloseste identitatea $\sum\frac {a}{b+c}=\frac 32+\frac 12\cdot\sum\frac {(b-c)^2}{(a+b)(a+c)}\ .$

Metoda 3. Folosim principiul rearanjamentelor $\implies\ \left|\begin{array}{c} \sum \frac {a}{b+c}\ge \sum\frac {b}{b+c}\\\\ \sum \frac {a}{b+c}\ge \sum\frac {c}{b+c}\end{array}\right|\bigoplus\ \implies$ inegalitatea Nesbitt.

Exemplu. In tr-un triunghi neobtuzunghic $ABC$ exista inegalitatea Walker $\boxed {\ a^2 + b^2 + c^2\ \ge\ 4(R + r)^2\ }$ .

Metoda 1. $\frac {a^2 + b^2 + c^2}{8RS} = \sum\frac {\cos^2A}{a\cdot\cos A}\ \stackrel{(C.B.S.)}{\ge}\ \frac {\left(\sum\cos A\right)^2}{\sum a\cdot\cos A} =$ $\frac {(R + r)^2}{2RS}\ \implies\ a^2 + b^2 + c^2\ \ge\ 4(R + r)^2$ .

Observatie. $HA = 2\cdot\delta_{BC}(O)$ si $\sum\delta_{BC}(O) = R + r$ (relatia Euler intr-un triunghi neobtuzunghic) $\implies\ \sum HA = 2(R + r)$

$\implies\ \boxed {\ \sum a^2\ge \left(\sum HA\right)^2\ }$ . Am notat $\delta_{BC}(O)$ - distanta de la circumcentrul $O$ al $\triangle ABC$ la dreapta $BC$ .

Metoda 2. $2 = \sum\frac {HA}{h_a} =$ $\sum\frac {HA^2}{h_a\cdot HA}\ \stackrel{(C.B.S.)}{\ge}\ \frac {\left(\sum HA\right)^2}{\sum h_a\cdot HA} =$ $\frac {\left(\sum HA\right)^2}{\frac 12\cdot\sum a^2}\ \implies\ \sum a^2\ge\ \left(\sum HA\right)^2$ .

Am folosit relatiile cunoscute $bc = 2Rh_a$ etc si $b^2 + c^2 - a^2 = 2bc\cdot\cos A$ etc. Vezi here articolul lui Cezar Lupu.

$\bigodot$ Extension of the Walker's inequality (see here). Let $M$ be an interior point of the triangle $ABC$ with the barycentrical coordinates

$(x,y,z)$ w.r.t. $\triangle ABC$ $(x + y + z = 1)$ . Then there is an inequality $\boxed {\ \left(MA + MB + MC\right)^2\ \le\ 2\cdot \sum \left(b^2z + c^2y - \frac {a^2yz}{y + z}\right)\ }$ .

Proof. For the point $M(x,y,z)$ denote $D\in BC\cap AM$ , $E\in CA\cap BM$ , $F\in AB\cap CM$ . Since $AM = (y + z)\cdot AD$

a.s.o. obtain that $\left(\sum MA\right)^2 =$ $\left[\sum (y + z)\cdot AD\right]^2\le \sum (y + z)\cdot\sum (y + z)\cdot AD^2 = 2\cdot\sum (y + z)\cdot AD^2 =$

$2\cdot\sum\left(b^2z + c^2y - \frac {a^2yz}{y + z}\right)$ . In conclusion $\left(\sum MA\right)^2\le 2\cdot$ $\sum\left(b^2z + c^2y - \frac {a^2yz}{y + z}\right)$ with equality iff $AD = BE = CF$ .

Particular cases.

$\odot\ M: = H\ \implies\ \left(\sum HA\right)^2\ \le\ a^2 + b^2 + c^2\ \le \ 3\cdot\sum HA^2$ .

$\odot\ M: = G\ \implies\ \left(\sum GA\right)^2\ \le\ a^2 + b^2 + c^2\ = \ 3\cdot\sum GA^2$ .

$\odot\ M: = I\ \implies\ \left(\sum IA\right)^2\ \le\ 2abc\cdot\sum\left(\frac 1a - \frac {1}{b + c}\right)\ \le\ a^2 +$ $b^2 + c^2\ \le\ \ 3\cdot\sum IA^2$ .

$\bigodot\ \ \left|\begin{array}{c} 0\le X\le A\\\\ 0\le Y\le B\\\\ 0\le P\le \sqrt {XY}\end{array}\right|\implies\left|\begin{array}{c} \underline {\underline {(A-X)(B-Y)}}\ \le\ \left(\sqrt {AB}-\sqrt {XY}\right)^2\ \le\ \underline {\underline {\left(\sqrt {AB}-P\right)}}^2\\\\ \sqrt {A-X}+\sqrt {B-Y}\ \le\ \sqrt {\left(\sqrt A+\sqrt B\right)^2-\left(\sqrt X+\sqrt Y\right)^2}\end{array}\right|$

Exemplu 1. Fie numerele reale $a_{k}$ , $b_k$ , unde $k\in\overline {1,n}$ si numerele reale pozitive $a$ , $b$ pentru care $a^2\ge \sum_{k=1}^na_k^2$ si $b^2\ge\sum _{k=1}^nb^2_k$ .

Atunci exista inegalitatea Aczel $\boxed {\ \left(a^2-\sum _{k=1}^na^2_k\right)\left(b^2-\sum _{k=1}^nb^2_k\right)\ \le\ \left(ab-\sum_{k=1}^na_kb_k\right)^2\ }\ .$ Aceasta inegalitate se obtine

imediat prin inlocuirile $X=\sum_{k=1}^na_k^2\le a^2=A\ ,\ Y=\sum_{k=1}^nb_k^2\le b^2=B$ si $P=\sum_{k=1}^na_kb_k\le \sqrt {XY}$ (inegalitatea C.B.S.).

O interesanta consecinta este $\sqrt {a^2-\sum_{k=1}^na^2_k}+\sqrt {b^2-\sum_{k=1}^nb^2_k}$ $\ \le\ \sqrt {(a+b)^2-\sum_{k=1}^n\left(a_k+b_k\right)^2}\ .$

Exemplu 2. Asemanator se pot genera asa zise " inegalitati de tip Aczel ". Pentru $\left|\ \begin{array}{c} 0 < a_1\le a_2\le \ldots \le a_n\\\\ x_1\ge x_2\ge \ldots\ge x_n>0\end{array}\ \right|$ , notam

$\begin{array}{c} X=\left(a_1+a_2+\ldots +a_n\right)^2\le n\cdot\left(a_1^2+a_2^2+\ldots +a_n^2\right)=A\\\\ Y=\left(x_1+x_2+\ldots +x_n\right)^2\le n\cdot\left(x_1^2+x_2^2+\ldots +x_n^2\right)=B\\\\ P=n\cdot\left(a_1x_1+a_2x_2+\ldots +a_nx_n\right)\le\left(a_1+a_2+\ldots +a_n\right)\cdot\left(x_1+x_2+\ldots +x_n\right)=\sqrt {XY}\end{array}$

Se obtine astfel lantul de inegalitati $\frac 1n\cdot\sum_{1\le j < k\le n}^n\left|\left(a_j - a_k\right)\left(x_j - x_k\right)\right| + \sum_{k = 1}^n a_kx_k\ \le$

$\frac 1n\cdot\sqrt {\sum_{1\le j<k\le n}^n (a_j-a_k)^2\cdot\sum_{1\le j<k\le n}^n (x_j-x_k)^2 + \sum_{k=1}^n a_kx_k}\ \le\ \sqrt {\sum_{k=1}^n a_k^2\cdot\sum_{k=1}^n x_k^2}$

care este o intarire a inegalitatii C.B.S. in situatia speciala in care cele doua siruri finite $a_k$ , $x_k$ , $k\in\overline {1,n}$ sunt invers ordonate.

Exemplu 3. Se stie inegalitatea bilaterala $3(ab+bc+ca)\le (a+b+c)^2\le 3\left(a^2+b^2+c^2\right)\ .$

$\odot\ \ \left|\begin{array}{c} X=3\cdot\sum bc\le \left(\sum a\right)^2=A\\\\ Y=\sum bc\le \sum a^2=B\\\\ P=\sqrt 3\cdot\sum bc=\sqrt {XY}\\\\ \left[\ A-X=B-Y=\frac 12\cdot\sum (b-c)^2\ \right]\end{array}\right|$ $\implies$ $(ab+bc+ca)\sqrt 3+\frac 12\cdot \sum (b-c)^2\le (a+b+c)\cdot\sqrt {a^2+b^2+c^2}\ .$

$\odot\ \ \left|\begin{array}{c} X=\left(\sum a\right)^2\le 3\cdot\sum a^2=A\\\\ Y=\sum bc\le \sum a^2=B\\\\ P=3\cdot\sqrt {abc\sum a}\le \sum a\cdot \sqrt {\sum bc}=\sqrt {XY}\\\\ \left[\ A-X=2(B-Y)=\sum (b-c)^2\ \right]\end{array}\right|$ $\implies$ $3\sqrt {abc(a+b+c)}+\frac {\sqrt 2}{2}\cdot \sum (b-c)^2\le \left(a^2+b^2+c^2\right)\sqrt 3\ .$

$\odot\ \ \left|\begin{array}{c} X=3\cdot\sum bc\le \left(\sum a\right)^2=A\\\\ Y=\left(\sum a\right)^2\le 3\cdot\sum a^2=B\\\\ P=3\cdot\sum bc\le \sqrt 3\cdot\sum a\cdot \sqrt {\sum a^2}=\sqrt {XY}\\\\ \left[\ 2(A-X)=(B-Y)=\sum (b-c)^2\ \right]\end{array}\right|$ $\implies$ $3(ab+bc+ca)+\frac {\sqrt 2}{2}\cdot \sum (b-c)^2\le (a+b+c)\cdot\sqrt {3\left(a^2+b^2+c^2\right)}\ .$

$\bigodot$ Fie numerele reale nenegative $x$ , $y$ , $z$ . Atunci pentru orice numar pozitiv $r$ exista relatia

$\boxed {\ x^r(x-y)(x-z)+y^r(y-x)(y-z)+z^r(z-x)(z-y)\ }\ \ge\ 0$ (inegalitatea Schur).

Demonstratie. Presupunem fara a restrange generalitatea ca $x\ge y\ge z\ .$ Se observa ca

$x\ge y\ge z\ \implies$ $\left|\begin{array}{c} x-z\ge y-z\ge 0\\\\ x^r\ge y^r\ge 0\end{array}\right|\ \implies\ x^r(x-z)\ge y^r(y-z)\ .$ In concluzie,

$\sum x^r(x-y)(x-z)=(x-y)\cdot\left[x^r(x-z)-y^r(y-z)\right]+z^r(x-z)(y-z)\ge 0\ .$

Extindere I. Daca tripletele $(a,b,c)$ , $(x,y,z)$ de numere reale sunt la fel ordonate, atunci $\sum a(x-y)(x-z)\ \ge\ 0\ .$

Extindere II. Fie numerele reale $a$ , $b$ , $c$ si $x$ , $y$ , $z$ astfel incat $a\ge b\ge c$ si $y\in [x,z]\cup [z,x]\ .$

Consideram o functie convexa sau monotona $f: \mathbb R\rightarrow [0,\infty )$ si $n\in \mathbb N^*\ .$ Atunci $\sum f(x)(a-b)^n(a-c)^n\ge 0\ .$

$\odot\ \sum a(a-b)(a-c)\ge 0\ \Longleftrightarrow\ 2\sum a^3+$ $3abc\ge \sum a\cdot\sum a^2\ \Longleftrightarrow\ \sum a^3+6abc\ge \sum a\cdot \sum bc\ .$

$\odot\ \sum a^2(a-b)(a-c)\ge 0\ \Longleftrightarrow\ 2\sum a^4+$ $abc\sum a\ge \sum a\cdot\sum a^3\ \Longleftrightarrow\ \sum a^4+2abc\sum a\ge \sum a^2\cdot\sum bc\ .$

$\bigodot\ \{x\ ,\ y\ ,\ z\}\ \subset\ R\ \wedge\ x+y+z=0$ $\Longrightarrow$ $\boxed { yza^2+zxb^2+xyc^2\ \le\ 0\ }\ .$

Demonstratie. $\left\|\ \begin{array}{c} yza^2+zxb^2+xyc^2\ \le\ 0\\\\ z=-(x+y)\end{array}\ \right\|$ $\Longrightarrow\ yza^2\le (y+z)\left(zb^2+yc^2\right)\ \Longleftrightarrow$

$y^2c^2+yz\left(b^2+c^2-a^2\right)+z^2b^2\ge 0$ $\Longleftrightarrow\ y^2c^2+2yzbc\cdot \cos A+z^2b^2\ge 0\ \Longleftrightarrow$

$\left(yc+zb\cdot\cos A\right)^2+z^2b^2\sin^2A\ge 0\ ,$ ceea ce este adevarat. Avem egalitate $\Longleftrightarrow\ x=y=z=0 .$

Cazuri particulare.

$\ \bullet\ x=b-c\ ,\ y=c-a\ ,\ z=a-b$ $\Longrightarrow\ \sum (c-a)(a-b)a^2\le0\ \Longrightarrow$

$\underline {\overline {\left\|\ a^2(a-b)(a-c)+b^2(b-c)(b-a)+c^2(c-a)(c-b)\ \ge\ 0\ \right\|}}$ (inegalitatea Schur pentru $r=2$ ).

$\ \bullet\ x=a(b-c)\ ,\ y=b(c-a)\ ,\ z=c(a-b)$ $\Longrightarrow\ \sum b(c-a)c(a-b)a^2\le0\ \Longrightarrow$

$\boxed { a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)\ \ge\ 0\ }$ (inegalitatea Schur pentru $r=1$ ).

Inegalitatea o regasim si in una din urmatoarele forme : $\left\|\begin{array}{c} (a+b+c)^3+9abc\ \ge\ 4(a+b+c)(ab+bc+ca)\ .\\\\ 2\left(a^3+b^3+c^3\right)+3abc\ \ge\ (a+b+c)\left(a^2+b^2+c^2\right)\ .\end{array}\right\|\ .$

$\ \bullet\ x=b+c-2a\ ,\ y=c+a-2b\ ,\ z=a+b-2c$ $\Longleftrightarrow$ $\sum (c+a-2b)(a+b-2c)a^2\le0\ \Longleftrightarrow$

$s_1^3-5s_1s_2+18s_3\ \le\ 0$ $\Longleftrightarrow$ $8p^3-10p\left(p^2+r^2+4Rr\right)+18\cdot 4Rpr\ \le\ 0$ $\Longleftrightarrow$ $\boxed { p^2+5r^2\ \ge\ 16Rr\ }\ .$

$\ \bullet\ x=\frac {a^2-bc}{(a+b)(a+c)}$ etc. Se observa ca $\sum\frac {a^2-bc}{(a+b)(a+c)}=0\ .$ Obtinem $\sum \frac {a(a+b)(a+c)}{b+c}\ \le\ \sum\frac {a^2\left(b^2+c^2\right)}{bc}\ .$

$\bigodot$ Fie un triunghi $ABC$ si notam proiectiile $X$ , $Y$ , $Z$ ale unui punct $P$ interior triunghiului pe dreptele suport $BC$ , $CA$ , $AB$

ale laturilor acestuia. Atunci $\boxed {\ PA+PB+PC\ \ge\ 2\cdot (PX+PY+PZ)\ }$ (inegalitatea Erdos-Mordell - vezi aici).

Demonstratie. Proiectam segmentul $[YZ]$ pe $BC\ \ : \ \ PA\cdot \sin A= YZ\ge \Pr_{BC}([YZ])=\Pr_{BC}([PY])+\Pr_{BC}([PZ])\implies$

$PA\cdot\sin A\ge PY\cdot\sin C+PZ\cdot \sin B$ $\Longrightarrow$ $a\cdot PA\ge c\cdot PY+b\cdot PZ$ . $\Longrightarrow$ $PA\ge \frac ca\cdot PY+\frac ba\cdot PZ$ . Obtinem :

$\left\|\begin{array}{c} PA\ge \frac ca\cdot PY+\frac ba\cdot PZ\\\\ PB\ge \frac ab\cdot PZ+\frac cb\cdot PX\\\\ PC\ge \frac bc\cdot PX+\frac ac\cdot PY\end{array}\right| \bigoplus\ \Longrightarrow$ $PA+PB+PC\ge \sum\left(\frac bc+\frac cb\right)\cdot PX\ge 2(PX+PY+PZ)\ .$

$\odot$ Extindere I. Fie un triunghi $ABC$ si punctele $D\in (BC)$ , $E\in (CA)$ , $F\in (AB)$ . Notam proiectiile

$X$ , $Y$ , $Z$ ale unui punct $P$ interior triunghiului pe dreptele suport $BC$ , $CA$ , $AB$ . Atunci exista inegalitatea

$\boxed {\ PA+PB+PC\ \ge\ 2\cdot\left(PX\sqrt {\frac {BF\cdot CE}{DF\cdot DE}}+PY\sqrt {\frac {CD\cdot AF}{ED\cdot EF}}+PZ\sqrt {\frac {AE\cdot BD}{FE\cdot FD}}\right)\ }\ .$

Demonstratie. Proiectam segmentele $[YZ]$ , $[ZX]$ , $[XY]$ pe $EF$ , $FD$ , $DE$ respectiv $: \ PA\cdot\sin A=YZ\ge\Pr_{EF}([YZ])=$

$\Pr_{EF}([PY])+\Pr_{EF}([PZ])=PY\cdot\sin\widehat {AEF}+PZ\cdot\sin\widehat {AFE}$ $\implies$ $PA\ge PY\cdot\frac {\sin\widehat {AEF}}{\sin A}+PZ\cdot\frac {\sin\widehat {AFE}}{\sin A}\implies$

$PA\ge PY\cdot\frac {AF}{EF}+PZ\cdot\frac {AE}{EF}\ .$ Asadar, $\left\|\begin{array}{c} PA\ge PY\cdot\frac {AF}{EF}+PZ\cdot\frac {AE}{EF}\\\\ PB\ge PZ\cdot\frac {BD}{FD}+PX\cdot\frac {BF}{FD}\\\\ PC\ge PX\cdot\frac {CE}{DE}+PY\cdot\frac {CD}{DE}\end{array}\right|\bigoplus\implies$

$\sum PA\ge\sum PX\cdot\left(\frac {BF}{FD}+\frac {CE}{DE}\right)$ $\implies$ $\sum PA\ge\sum 2\cdot PX\cdot\sqrt {\frac {BF\cdot CE}{DF\cdot DE}}\ .$

Cazuri particulare. Daca $D$ , $E$ , $F$ sunt mijloacele laturilor corespunzatoare, atunci regasim inegalitatea Erdos-Mordell.

Daca $D$ , $E$ , $F$ sunt proiectiile lui $H$ pe laturile corespunzatoare se obtine inegalitatea $\boxed {\ \sum PA\ \ge\ 2\cdot\sum \frac {a}{\sqrt {bc}}\cdot PX\ }\ .$

Demonstratie directa : $\sum a\cdot PX=2S\le a\cdot AX\le a\cdot (PA+PX)\implies$ $PA\ge \frac ba\cdot PY+\frac ca\cdot PZ\ .$

In concluzie, $\left\|\begin{array}{c} PA\ge \frac ba\cdot PY+\frac ca\cdot PZ\\\\ PB\ge \frac cb\cdot PZ+\frac ab\cdot PX\\\\ PC\ge \frac ac\cdot PX+\frac bc\cdot PY\end{array}\right|\bigoplus$ $\implies$ $\sum PA\ge \sum \left(\frac ab+\frac ac\right)\cdot PX\ge$ $2\cdot\sum \frac {a}{\sqrt {bc}}\cdot PX\ .$

$\odot$ Extindere II. Fie un triunghi $ABC$ si punctele $M$ , $N$ , $P$ interioare triunghiului $ABC$ astfel incat $NP \perp BC$ ,

$PM \perp CA$ si $MN \perp AB$ . Notam $X\in NP\cap BC$ , $Y\in PM\cap CA$ si $Z\in MN\cap AB$ . Atunci exista inegalitatea

$\boxed {\ MA\ +\ NB\ +\ PC\ \ge\ MY\ +\ MZ\ +\ NZ\ +\ NX\ +\ PX\ +\ PY\ }\ .$

Demonstratie. Se observa ca $ABC\sim MNP$ , adica $\frac {a}{NP}=\frac {b}{PM}=\frac {c}{MN}$ . Asadar, $\sum\left(NZ-MZ\right)\left(\frac{a}{b}-\frac{b}{a}\right)=$

$\epsilon\cdot\sum MN\left(\frac{a}{b}-\frac{b}{a}\right)=\epsilon\sum c\left(\frac{a}{b}-\frac{b}{a}\right)=\frac {\epsilon}{abc}\cdot\sum c^2\left(a^2-b^2\right)=0$ $\implies$ $\sum\left(NZ-MZ\right)\left(\frac{a}{b}-\frac{b}{a}\right)=0\ (*)\ ,$

unde $\epsilon^2=1\ .$ Se arata usor ca $MA \geq MY \cdot \frac{c}{a} + MZ \cdot \frac{b}{a}$ . Similar $\left\|\begin{array}{c} MA \geq MY \cdot \frac{c}{a} + MZ \cdot \frac{b}{a}\\\\ NB \geq NZ \cdot \frac{a}{b} + NX \cdot \frac{c}{b}\\\\ PC \geq PX \cdot \frac{b}{c} + PY \cdot \frac{a}{c}\end{array}\right|\bigoplus$ $\implies$

$MA+NB+PC \ge\sum \left(MZ \cdot \frac{b}{a} + NZ \cdot \frac{a}{b}\right)=$ $\sum \left[\frac{1}{2}\left(NZ-MZ\right)\left(\frac{a}{b}-\frac{b}{a}\right) + \frac{1}{2}\left(NZ+MZ\right)\left(\frac{a}{b}+\frac{b}{a}\right)\right]\stackrel{(*)}{\ =\ }$

$\sum\frac{1}{2}\left[\left(NZ+MZ\right)\left(\frac{a}{b}+\frac{b}{a}\right)\right]\ge$ $\sum (NZ+MZ)$ $\implies$ $MA+NB+PC\ge MY+MZ+NZ+NX+PX+PY\ .$

$\bigodot\ \boxed {\ 3r\sqrt 3\le p\le\frac {3R\sqrt 3}{2}\ }$ (Mitrinovic) ; $\boxed {\ p\sqrt 3\ \le\ 4R+r\ }\ \ ;\ \ \boxed {\ 3r(4R+r)\ \le\ p^2\ }\ \ ;\ \ \boxed {\ p-3r\sqrt 3\le 2(R-2r)\ }$ (Blundon).

$\ \boxed {\ a^2+b^2+c^2\ \le\ 4\left(2R^2+r^2\right)\ \le\ 9R^2\ }\ ;\ \boxed {\ 16Rr-5r^2\ \le\ p^2\ \le\ \frac {R(4R+r)^2}{2(2R-r)}\ \le\ 4R^2+4Rr+3r^2\ }$ (Blundon & Gerretsen).

Demonstratie.

$\odot$ Se stie ca $: \ abc=4Rpr\ ;\ (p-a)(p-b)(p-c)=pr^2\ ;\ \sum (p-b)(p-c)=r(4R+r)\ .$

Din produsul inegalitatilor evidente $a^2\ge a^2-(b-c)^2$ , adica $a^2\ge 4(p-b)(p-c)$ etc., $\Longrightarrow$

$\ abc\ge 8(p-a)(p-b)(p-c)$ , adica $4Rpr\ge 8pr^2$ de unde obtinem $\underline {R\ge 2r}$ . Aplicam $\mathrm {A.M.\ \ge\ G.M.}$ :

$\ \frac {(p-a)+(p-b)+(p-c)}{3}\ge$ $\sqrt[3]{p-a)(p-b)(p-c)}$ $\Longrightarrow$ $p\ge 3\sqrt[3]{pr^2}$ $\Longrightarrow$ $p^2\ge 27r^2\Longrightarrow$ $\underline {p\ge 3r\sqrt 3}\ .$

Este necesar sa aratam acum ca $2p\le 3R\sqrt 3\ .$ In inegalitatea $3(xy+yz+zx)\le (x+y+z)^2$ inlocuim

$\left\| \begin{array}{c} x=(p-b)(p-c)\\\\ y=(p-c)(p-a)\\\\ z=(p-a)(p-b)\end{array}\ \right\|\ \Longrightarrow$ $3p(p-a)(p-b)(p-c)\le \left[r(4R+r)\right]^2$ $\Longrightarrow$ $\underline {p\sqrt 3\le 4R+r}\ .$

Adunam relatiile $9p\sqrt 3\le 9(4R+r)$ si $9r\le p\sqrt 3\ \Longrightarrow\ 8p\sqrt 3\le 36R$ $\Longrightarrow$ $\underline {2p\le 3R\sqrt 3}$ (vezi si aici).

Remark. I"ll show that $\frac pR=\sin A+\sin B+\sin C\le 3\cdot \sin \frac {A+B+C}{3}=\frac {3\sqrt 3}{2}$ . Observe that for $0<x\le y<\pi$ have

$\sin x+\sin y=2\sin \frac {x+y}{2}\cos\frac {x-y}{2}\le 2\sin\frac {x+y}{2}$ , i.e. $\boxed{0<x\le y<\pi\implies\sin x+\sin y\le \sin\frac {x+y}{2}}$ . Therefore,

$\left\{\begin{array}{c} \sin A+\sin B\le 2\sin \frac{A+B}{2}\\\\ \sin C+\sin \frac{\pi}{3} \le 2\sin \frac{C+\frac{\pi}{3}}{2}\\\\ \sin \frac{A+B}{2}+\sin \frac{C+\frac{\pi}{3}}{2} \le 2 \sin \frac{A+B+C+\frac{\pi}{3}}{4}=2\sin \frac{\pi}{3}\end{array}\right\|\ \implies\ \sum$ $\sin A+\sin \frac{\pi}{3} \le 4\sin \frac{\pi}{3}$ $\implies\sum\sin A\le 3\sin \frac{\pi}{3} =$ $\frac{3\sqrt{3}}{2}$ .

$\odot$ Din inegalitatea $3\cdot\sum (p-b)(p-c)\ \le \left[\sum (p-a)\right]^2$ obtinem $\underline {3r(4R+r)\ \le p^2}$ , deoarece

$\sum (p-b)(p-c)=r(4R+r)\ .\ OG^2=R^2+p_w(G)=R^2-\frac {a^2+b^2+c^2}{9}\ge 0$ $\Longrightarrow$ $\boxed {\ a^2+b^2+c^2\ \le\ 9R^2\ }\ .$

In concluzie, am dovedit (intr-o exprimare compacta !) ca $\boxed {\ 1\ \le\ \frac {4R+r}{p\sqrt 3}\ \le\ \frac {p}{3r\sqrt 3}\ \le\ \frac {R}{2r}\ }\ .$

$\odot$ Tinand seama ca $IH^2=4R^2+4Rr+3r^2-p^2\ge 0$ obtinem inegalitatea $\boxed {\ p^2\ \le\ 4R^2+4Rr+3r^2\ }\ .$ Asadar,

$\sum a^2=2\cdot\left[p^2-r(4R+r)\right]\ \le\ 2\left[\left(4R^2+4Rr+3r^2\right)-r(4R+r)\right]$ $\Longrightarrow$ $\boxed {\sum a^2\ \le\ 4\left(2R^2+r^2\right)\ \le\ 9R^2\ }\ .$

$\odot\ \left\|\ \begin{array}{c} IH^2=4R^2+4Rr+3r^2-p^2\ge 0\\\\ 9\cdot IG^2=IN^2=p^2+5r^2-16Rr\ge 0\\\\ H\Gamma^2=4R^2\left[1-\frac {2p^2(2R-r)}{R(4R+r)^2}\right]\ge 0\end{array}\ \right\|$ $\Longrightarrow$ $\boxed {\ \begin{array}{c} 16Rr-5r^2\ \le\ p^2\ \le\ 4R^2+4Rr+3r^2\\\\ 2p^2(2R-r)\ \le\ R(4R+r)^2\end{array}\ }\ .$

$\odot$ Se stie ca $p^2\le 4R^2+4Rr+3r^2$ si se arata usor ca $4R^2+4Rr+3r^2\le \left[2R+\left(3\sqrt 3-4\right)r\right]^2\ .$

In concluzie, $p^2\le \left[2R+\left(3\sqrt 3-4\right)r\right]^2$ $\Longleftrightarrow$ $p\le 2R+\left(3\sqrt 3-4\right)r$ , adica $\boxed {\ p-3r\sqrt 3\le 2(R-2r)\ }\ .$

Observatii.

$\odot$ Pentru un triunghi $ABC$ notam $\left\|\ \begin{array}{c} A=(b+c)(c+a)(a+b)-8abc\\\\ B=abc-(b+c-a)(c+a-b)(a+b-c)\end{array}\ \right\|$ . Atunci $A\ge B\ge 0$ , adica

inegalitatea $\prod (b+c)+\prod (b+c-a)\ge 9abc$ (Virgil Nicula & Cosmin Pohoata, Mathematical Reflections).

Intr-adevar, $A=\sum a\cdot\sum bc-9abc=2p(p^2+r^2+4Rpr)-36Rpr=2p(p^2+r^2-14Rr)\ge 0$ deoarece

$p^2\ge 16Rr-5r^2\ge 14Rr-r^2\ .$ Pe de alta parte, $B=abc-8\prod (p-a)=4Rpr-8pr^2=4pr(R-2r)\ge 0$ .

In sfarsit, $A-B=2p(p^2+r^2-14Rr-2Rr+4r^2)=2p(p^2+5r^2-16Rr)\ge 0$ .

$\odot\ \ OH^2=9R^2-2p^2+2r(4R+r)\ge 0\ \Longrightarrow$ $\left\|\begin{array}{c} 2\left(p^2-27r^2\right)\le(R-2r)\left(9R+26r\right)\\\\ (R-2r)(9R+2r)\le\ 27R^2-4p^2\end{array}\ \right\|\ .$

Obtinem lantul de implicatii $3r\sqrt 3\ \le\ p\ \implies\ 2r\le\ R\ \implies\ 2p\ \le\ 3R\sqrt 3\ \ (1)\ .$

$IN^2=p^2+5r^2-16R^2\ge0\ \implies$ $\left\|\begin{array}{c} 16r(R-2r)\le p^2-27r^2\\\\ 27R^2-4p^2\le (R-2r)(27R-10r)\end{array}\right\|\ .$

Obtinem lantul de implicatii $2p\ \le\ 3R\sqrt 3\ \implies\ 2r\ \le\ R\ \implies\ 3r\sqrt3\ \le\ p\ \ (2)\ .$

Din sirurile de implicatii $(1)$ si $(2)$ se obtine lantul de echivalente $\boxed {\ 3r\sqrt 3\ \le\ p \Longleftrightarrow\ 2r\ \le\ R\ \Longleftrightarrow\ 2p\ \le\ 3R\sqrt 3\ }$

care exprima echivalenta intre cele trei inegalitati fundamentale din geometria triunghiului.

$\ \odot$ Din cele prezentate pana acum rezulta un lant interesant de inegalitati centrat in $p^2$ care este des folosit in aplicatii.

$27r^2\ \le\ 3r(4R+r)\ \le\ \frac {27Rr}{2}\ \le$ $\boxed {16Rr-5r^2\ \le\ p^2\ \le\ \frac {R(4R+r)^2}{2(2R-r)}}\ \le$ $\ 4R^2+4Rr+3r^2\ \le\ \frac {(4R+r)^3}{16R-5r}\ .$

Lantul poate fi "slabit" spre dreapta : $\frac {(4R+r)^3}{16R-5r}\ \le$ $\ \frac {(4R+r)^2}{3}\ \le\ \frac {R(4R+r)^3}{2(2R-r)(2R+5r)}\ \le$ $\ \frac {R(4R+r)^2}{2(R+r)}\ .$

$\odot\ 12r(2R-r)\ \le\ \sum a^2\ \le 4$ $\left(2R^2+r^2\right)\ \Longleftrightarrow\ 16Rr-5r^2\ \le\ p^2\ \le\ 4R^2+4Rr+3r^2\ .$

$\odot\ \left\|\ \begin{array}{c} r_a+r_b+r_c=4R+r\\\\ r_ar_b+r_br_c+r_cr_a=p^2\end{array}\ \right\|\ \ \wedge\ \ 3(r_ar_b+r_br_c+r_cr_a)\ \le\ (r_a+r_b+r_c)^2\ \Longrightarrow$ $\boxed {\ p\sqrt 3\ \le\ 4R+r\ }\ .$

$\odot\ NI^2+HI^2-HN^2=$ $\left(p^2+5r^2-16Rr\right)+\left(4R^2+4Rr+3r^2-p^2\right)-4R\left( R-2r\right)=-4r(R-2r)\le 0$ $\implies$

$\boxed {\ M(\widehat {GIH})\ge 90^{\circ}\ \Longleftrightarrow\ R\ge 2r\ }$ deoarece $2\cdot NI\cdot HI\cdot\cos\widehat {NIH}=NI^2+HI^2-HN^2\ \le\ 0$ si $G\in NI\ .$

Se poate usor arata ca intr-un triunghi neechilateral $ABC$ avem relatia $\sum IA^2\ <\ \sum HA^2\ .$ Din relatia

$\sum XA^2=3\cdot XG^2+\frac {a^2+b^2+c^2}{3}$ in particular pentru $X\in\{\ I\ ,\ H\ \}$ obtinem : $\left\|\ \begin{array}{c} \sum IA^2=3\cdot IG^2+\frac {a^2+b^2+c^2}{3}\\\\ \sum HA^2=3\cdot HG^2+\frac {a^2+b^2+c^2}{3}\end{array}\ \right\|$

Asadar, $\sum IA^2\ <\ \sum HA^2\ \Longleftrightarrow$ $IG\ <\ HG$ , ceea ce este adevarat deoarece $m(\angle GIH)> 90^{\circ} .$

Guinand's theorem. Daca $M$ este mijlocul lui $[HG]$ , atunci $IM\le OG$ deoarece $m(\angle GIH)\ge 90^{\circ}\iff$ $IM\le \frac 12\cdot HG=OG$ .

$\odot\ \boxed {\ OH^2\ \ge\ IO^2+2\cdot IH^2\ }\ .$ Intr-adevar, stiind ca $\left\|\ \begin{array}{c} OH^2=9R^2+8Rr+2r^2-2p^2\\\\ IO^2=R^2-2Rr\\\\ IH^2=4R^2+4Rr+3r^2-p^2\end{array}\ \right\|$

se arata usor ca inegalitatea propusa devine echivalenta cu inegalitatea remarcabila $R\ge 2r\ .$

See LXVII-message and here

This post has been edited 40 times. Last edited by Virgil Nicula, Apr 25, 2016, 3:32 AM

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Excellent blog!!! Congratulations!!! Gretting of Peru

by viterick, Nov 23, 2013, 3:13 PM

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170. A nice "slicing" proposed problem for middle school.

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148. A special class of systems.

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145. Spain MO P5 - Point on median.

144. A conditioned concurrency in a triangle.

143. Assess of a ratio (nice and difficult problem).

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138. Generalized "slicing" problems (own).

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134. Some problems with projections/perpendiculars (2).

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132. Asupra unei inegalitati intr-un triunghi.

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119. Inspired by a Miculita's problem.

118. An usual remarkable geometrical locus.

117. Some applications of harmonic division/quadrilateral.

116. The length of AE (nine methods !).

115. An difficult equation with one radical.

114. A property of tangential triangle for ABC.

113. Range for a function.

112. The equivalence relation ".s.s."

111. Simple, but interesting ...

110. German TST 2004 - Exam V , Problem 1

109. Two equal areas ==> find A.

108. Old, short and nice proof of Euler's relation.

107. XY parallel BC and X,Y are isogonal/izotomic points.

106. OLM - Bucuresti (geometrie).

105. A problem with three circles.

104. Concursul Revistei de Matematica din Galati (RMG).

103. Metrical usual relations in triangle. Applications.

102. Characterization of a metrical relation in a triangle.

101. O problema cu numere complexe.

100. Bobiller's theorem.

99. Some simple and nice geometry problems.

98. Symmedian & median (metrical relations).

97. Problem of butterfly.

August 2010

96. Integrals.

95. Three circles in an angle.

94. Equations of angle bisectors (interior and exterior).

93. The Appolonius' construct problems.

92. CA , CB tangent to parabola - OIMU 2008 Problem 4.

91. JBMO 2010, Problem 3.

90. Nice and difficult inequality in ABC (own).

89. Similarity (parallel/antiparallel).

88. A nice geometric locus conditioned metrically.

87. An interesting vectorial identity.

86. A very nice maxmin.

85. Problem "slicing" : 60-24-12.

84. IRAN national math olympiad - 2010

83. A metrical characterization of concurrence.

82. AA' , BN , CM concur in Gergonne point.

81. Sawayama and Thebault’s theorem.

80. Three concurrent perpendiculars (the Carnot's lemma).

79. A quadrilateral with many perpendicularities.

78. Jakobi's theorem.

77. Two equivalent perpendicularities.

76. A parallelogram and a circle (nice !).

75. Extended Steinbart Theorem.

74. Inegalitatea Erdos-Mordell.

73. A nice problems with a square and a perpendicularity.

July 2010

72. Colinearity in a triangle - H , P , M .

70. Colinearity with pole/polar - H\in PQ .

71. An angle questions (synthetic/trigonometric proofs).

69*1. Slicing problem 3.

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).

Submit

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by mathMagicOPS, Jan 9, 2025, 3:40 AM
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391345 views moment
by ryanbear, May 9, 2023, 6:10 AM
We need virgil nicula to return to aops, this blog is top 10 all time.
by OlympusHero, Sep 14, 2022, 4:44 AM
blog
by tigerzhang, Aug 1, 2021, 12:02 AM
Amazing blog.
by OlympusHero, May 13, 2021, 10:23 PM
the visits tho
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Bro this blog is ripped
by samrocksnature, Apr 14, 2021, 5:16 AM
Holy- Darn this is good. shame it's inactive now
by the_mathmagician, Jan 17, 2021, 7:43 PM
godly blog. opopop
by OlympusHero, Dec 30, 2020, 6:08 PM
long blog
by MrMustache, Nov 11, 2020, 4:52 PM
372554 views!
by mrmath0720, Sep 28, 2020, 1:11 AM
wow... i am lost.

369302 views!

-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
impressive
awesome. 358,000 visits?????
by OlympusHero, May 14, 2020, 8:43 PM
Nice blog.
The element of Mathematics is dense here.
I love the style of your posts.
Thanks for writing the blog! I can learn Geometry from here too.
by epsilonist, Jun 27, 2011, 6:00 AM
Thank you very much for this blog.
by Affine, May 7, 2011, 6:20 PM
You seem to have solved every existing problem.
by Goutham, May 2, 2011, 10:59 AM
I love your posts, man (those which I understand)! It's interesting how you punish integrals with recurrence relations and sometimes make seemingly unrelated integrals evaluate each other!
by Caspar, Apr 11, 2011, 2:36 PM
Thank you a lot dear Virgil for an interesting and instructive blog.
by vittasko, Mar 12, 2011, 9:42 AM
Interesting Blog
by ShahinBJK, Mar 12, 2011, 5:07 AM
Interesant, d-le Nicula.
by silent_control, Mar 3, 2011, 6:00 PM
Yours is the largest blog full of math equations as far as I have known! Congrats!
by Goutham, Jan 30, 2011, 2:06 PM
15001th view.
by fries4guys, Jan 20, 2011, 4:57 PM
Thanks to all !
by Virgil Nicula, Dec 11, 2010, 8:32 PM
El_Ectric:
You can either copy some parts from the blog post, paste it at "Search Blogs" (which is lower on the page, after "Blog Stats") and find the whole post.

You're welcome.
by Thomas_, Dec 8, 2010, 6:42 PM
Thanks Thomas_.
by El_Ectric, Dec 8, 2010, 4:21 PM
El_Ectric: Try to open this blog with Internet Explorer (browser). If it still doesn't work, just copy the post you're interested in, and than paste it into Word Document, or almost any kind of reading documents and you'll have all the details.

Nice blog!
by Thomas_, Nov 24, 2010, 4:59 PM
Dear Virgil !

Congratulations for this blog !
Babis stergiou - Greece
by stergiu, Nov 12, 2010, 6:15 AM

72 shouts

My (math)blog

4. Remarkable geometrical inequalities (2).

by Virgil Nicula, Apr 19, 2010, 2:22 PM

1 Comment