285. Nice and easy inequalities. For example, IMAC - 2009.

by Virgil Nicula, Jun 13, 2011, 1:42 PM

IMAC - 2009. Prove that for the sides of triangle $ABC$ exists the inequality $\frac{a^3}{b+c-a} + \frac{b^3}{a+c-b} + \frac{c^3}{a+b-c} \geq a^2 + b^2 + c^2$ .

Proof 1. $\sum\, \frac {a^3}{b+c-a}=\sum\, \frac {a^4}{a(b+c-a)}\, \stackrel{C.B.S.}{\ge}\, \frac {\left(a^2+b^2+c^2\right)^2}{2(ab+bc+ca)-(a^2+b^2+c^2)}$ thus remaining to show that

$\frac {\left(a^2+b^2+c^2\right)^2}{2(ab+bc+ca)-(a^2+b^2+c^2)}\, \ge\, a^2+b^2+c^2\iff a^2+b^2+c^2\ge ab+bc+ca$ , which is obviously.

Proof 2. Since $\sum (b+c-a)=\sum a$ and $a^2\le b^2\le c^2\iff$ $a\le b\le c\iff$ $\frac {1}{b+c-a}\le\frac {1}{c+a-b}\le\frac {1}{a+b-c}\iff$ $a^3\le b^3\le c^3$ .

Can apply the Chebyshev's inequality so that $:\ \left\{\begin{array}{c} \sum a^2\cdot\sum a\le 3\cdot\sum a^3\\\\ 9\le \sum (b+c-a)\cdot\sum \frac {1}{b+c-a}\\\\ \sum a^3\cdot \sum\frac {1}{b+c-a}\le 3\cdot\sum\frac {a^3}{b+c-a}\end{array}\right\|\ \bigodot\ \implies\ \sum a^2\le \sum\frac {a^3}{b+c-a}$ .

Similar proposed problem. Prove that $\sum\frac {a^3}{(s-b)(s-c)}\ge \frac {24\cdot S}{2R-r}$ and $\sum\frac {1}{s-a}\ge \frac {s}{r(2R-r)}$ , where $S=[ABC]$ and $2s=a+b+c$ .

Proof. I"ll use the well-known identity $\sum a(s-b)(s-c)=2sr(2R-r)$ . Thus, $\sum\frac {a^3}{(s-b)(s-c)}=$ $\sum\frac {\left(a^2\right)^2}{a(s-b)(s-c)}\stackrel{(C.B.S.)}{\ge}$

$\frac {\left(\sum a^2\right)^2}{\sum a(s-b)(s-c)}=$ $\frac {\left(\sum a^2\right)^2}{2sr(2R-r)}\stackrel{(\mathrm{Weitzenbock})}{\ge}$ $\frac {48\cdot S^2}{2\cdot S(2R-r)}=$ $\frac {24\cdot S}{2R-r}$ . I"ll proceed analogously $:\ \sum\frac {1}{s-a}=$ $\sum \frac {(s-b)+(s-c)}{2(s-b)(s-c)}=$

$\sum \frac {a}{2(s-b)(s-c)}=$ $\sum\frac {a^2}{2a(s-b)(s-c)}\stackrel{(C.B.S.)}{\ge}$ $\frac {(\sum a)^2}{2\cdot \sum a(s-b)(s-c)}=$ $\frac {2s^2}{2sr(2R-r)}=\frac {s}{r(2R-r)}$ .

Otherwise. $\sum\frac {1}{s-a}=$ $\frac {\sum (s-b)(s-c)}{\prod (s-a)}=$ $\frac {r(4R+r)}{sr^2}=$ $\frac {4R+r}{sr}$ . Therefore, $\sum\frac {1}{s-a}\ge \frac {s}{r(2R-r)}\iff$ $(4R+r)(2R-r)\ge s^2$ .

Observe that $s^2\stackrel{\mathrm{(Gerretsen)}}{\le} 4R^2+4Rr+3r^2$ and $4R^2+4Rr+3r^2\le (4R+r)(2R-r)\iff$ $2R^2-3Rr-2r^2\ge 0\iff$

$(R-2r)(2R+r)\ge 0$ what is obviously. Therefore obtain that $s^2\le (4R+r)(2R-r)$ . In conclusion and $\sum\frac {1}{s-a}\ge \frac {s}{r(2R-r)}$ .

PP2. Prove that $\left\{a,b,c\right\}\subset (0,1]\ \implies\ a+b+c+ \frac{1}{abc} \ge\frac{1}{a} + \frac{1}{b} + \frac{1}{c} +abc$ .

Proof. Denote $\left\{x,y,z\right\}\subset (0,1]$ so that $ax=by=cz=abc\equiv p$ . Observe that $xyz=p^2$ and our inequality

becomes $p\cdot\sum \frac 1x+\frac 1p\ge\frac 1p\cdot\sum x+p\iff$ $\frac 1p\cdot \sum yz+\frac 1p\ge \frac 1p\cdot\sum x+\frac 1p\cdot xyz\iff$ $1-\sum x\ +$

$\sum yz-xyz\ge 0\iff$ $(1-x)(1-y)(1-z)\ge 0\iff$ $(1-bc)(1-ca)(1-ab)\ge 0$ , what is truly.

An easy and nice extension. Prove that $(\forall )\ k\in\overline{1,n}\ ,\ x_k\in (0,1]\implies \sum_{k=1}^nx_k+$ $\frac {1}{x_1x_2\ldots x_{n-1}x_n}\ \ge\ \sum _{k=1}^n\frac {1}{x_k}+$ $x_1x_2\ldots x_{n-1}x_n$ .

Proof. Denote $f(x)=\frac 1x-x$ , where $x\in (0,1]$ . Prove easily that $f(ab)\ge f(a)+f(b)\iff$ $(1-a)(1-b)(1-ab)\ge 0$ .

Therefore, $\left\{\begin{array}{c} f(x_1x_2)\ge f(x_1)+f(x_2)\\\\ f(x_1x_2x_3)\ge f(x_1x_2)+f(x_3)\\\\ f(x_1x_2x_3x_4)\ge f(x_1x_2x_3)+f(x_4)\\\ ...............................................................\\\ f(x_1x_2\ldots x_{n-1}x_n)\ge f(x_1x_2\ldots x_{n-1})+f(x_n)\end{array}\right|\ \bigoplus\ \implies\ f(x_1x_2\ldots x_n)$ $\ge \sum_{k=1}^nf(x_k)\iff$

$\frac {1}{x_1x_2\ldots x_n}-x_1x_2\ldots x_n\ge \sum_{k=1}^n\left(\frac {1}{x_k}-x_k\right)\iff$ $\sum_{k=1}^n x_k+\frac {1}{x_1x_2\ldots x_n}\ge \sum_{k=1}^n\frac {1}{x_k}+x_1x_2\ldots x_n$ .

PP3. Prove that for any point $P$ which is interior to $\triangle ABC$ there is the relation $\frac {PA}{bc}+\frac {PB}{ca}+\frac {PC}{ab}\ge\frac 1R$ .

Proof. Observe that $PA+\delta_{BC}(P)\ge h_a$ a.s.o. and $\sum a\cdot\delta_{BC}(P) =2S$ . Therefore,

$\sum a\cdot\left[PA+\delta_{BC}(P)\right]\ge\sum ah_a=6S\implies$ $\sum a\cdot PA\ge 4S\implies$ $\sum\frac {PA}{bc}\ge \frac {4S}{abc}=\frac 1R$ .

PP4. Find the minimum value of $E(x,y)\equiv \cos x+\cos y+2\cos (x+y)$ .

Proof. Denote $\left\{\begin{array}{c} x=a+b\\\ y=a-b\end{array}\right|$ . Observe that $E(x,y)=\cos (a+b)+\cos (a-b)+2\cos 2a=$ $2\cos a\cos b+2\left(2\cos^2a-1\right)=$

$4\underline{\cos ^2a}+2\cos b\cdot\underline{\cos a}-2=$ $\left(2\cos a+\frac 12\cdot\cos b\right)^2+\frac 14\cdot\sin^2 b-\frac 14-2=$ $\frac 14\cdot\left[\left(4\cos a+\cos b\right)^2+\sin^2b-9\right]\ge -\frac 94$ .

We have equality if and only if $4\cos a+\cos b=0\ \ \wedge\ \ \sin b=0$ , i.e. $\sin b=0\ \ \wedge\ \ \cos a=\frac {\epsilon}{4}$ , where $\epsilon^2=1$ .

PP5. Let $ABC$ be a triangle so that $b\ge c$ . Prove that $\frac ba+\frac {a}{b+c}\ge\sqrt {2+\frac {b^2-c^2}{a^2}}\ (*)$ with equality

iff $A=2C$ . Remark. If $B=90^{\circ}$ , then our inequality becomes $\frac ba+\frac {a}{b+c}\ge\sqrt 3$ with equality iff $A=60^{\circ}$ .

Proof 1. If $b=c$ , then the inequality $(*)$ becomes $\frac ba+\frac {a}{2b}\ge \sqrt 2\iff$ $(a-b\sqrt 2)^2\ge 0$ , what is true. We have equality iff $a=b\sqrt 2$ , i.e. $\sin \frac A2=\frac {\sqrt 2}{2}\iff$

$A=2C$ . Suppose $b>c$ and denote $b^2-c^2=\lambda a^2$ , i.e. $b^2=c^2+\lambda a^2$ , where $\lambda >0$ . Inequality $(*)$ becomes $\frac ba+\frac {b-c}{\lambda a}\ge \sqrt {2+\lambda}\iff$

$(1+\lambda )b\ge c+\lambda a\cdot\sqrt {2+\lambda }\iff$ $(\lambda +1)^2\left(\lambda a^2+c^2\right)\ge$ $c^2+2ac\lambda \sqrt {2+\lambda}+\lambda^2a^2(2+\lambda )$ $\iff$ $a^2-2ac\sqrt{2+\lambda }+(2+\lambda )c^2\ge 0\iff$

$\left(a-c\sqrt {2+\lambda }\right)^2\ge 0$ , what is true. We have equality iff $a=c\cdot\sqrt {2+\lambda}$ , i.e. $\left|a^2-c^2\right|=bc\iff$ $|A-B|=C\iff A=2B$ .

Remark. If $\lambda =1$ , i.e. $b^2=a^2+c^2 \iff$ $B=90^{\circ}$ , then $(*)$ becomes $\frac ba+\frac {a}{b+c}\ge \sqrt 3$ , with equality iff $A=60^{\circ}$ .

Proof 2. By squaring both sides of the given inequality one obtains: $\frac {b^2}{a^2}+\frac {a^2}{(b+c)^2}+\frac {2b}{b+c}\, \ge\, 2+\frac {b^2}{a^2}-\frac {c^2}{a^2}$ $\iff$ $\frac {a^2}{(b+c)^2}+\frac {c^2}{a^2}\, \ge\, \frac {2c}{b+c}$ , which

is trivial by AM-GM inequality. Equality occurs iff $\frac a{b+c}=\frac ca\iff$ $a^2=bc+c^2\iff$ $a^2+b^2-c^2=$ $b^2+bc\iff 2ab\cos C=b^2+bc\iff$

$2a\cos C=b+c\iff$ $2a\cos C=\frac {a^2}c\iff$ $2c\cos C=a\iff 2\sin C\cos C=$ $\sin A\iff\sin 2C=\sin A\iff A=2C$ .

PP6. Let $P$ be a point inside the $\triangle ABC$ . Denote the distances $x$ , $y$ and $z$ from $P$ to the sides of $\triangle ABC$ . Prove

that $R \leq \frac{a^2+b^2+c^2}{18\cdot\sqrt [3]{xyz}}$ , where R is the circumradius of the triangle. When does equality occur ? (Taiwan TST 2005)

Proof. $\sum ax=2S\implies$ $3\cdot \sqrt[3]{abcxyz}\le 2S\ (*)$ . So, $3\cdot\sqrt [3]{a^2b^2c^2}\le \sum a^2\implies$ $12RS=3\cdot a^2b^2c^2\le\sqrt [3]{abc}\cdot\sum a^2\implies$

$12RS\cdot \sqrt [3]{xyz}\le \left(a^2+b^2+c^2\right)\cdot \sqrt[3]{abcxyz}\stackrel{(*)}{\le} \left(a^2+b^2+c^2\right)\cdot\frac {2S}{3}\implies$ $18R\cdot\sqrt [3]{xyz}\le a^2+b^2+c^2$ .

PP7. Prove that $\{x,y,z\}\subset\mathbb R\ \implies\ \sqrt{x^2-xy+y^2}+\sqrt{y^2-yz+z^2}\ge \sqrt{x^2+xz+z^2}$ .

Proof 1. Minkowski's inequality : $\sqrt {{{\left( {y - \frac{x}{2}} \right)}^2} + {{\left( {\frac{{x\sqrt 3 }}{2}} \right)}^2}} + \sqrt {{{\left( {\frac{z}{2} - y} \right)}^2} + {{\left( {\frac{{z\sqrt 3 }}{2}} \right)}^2}}$ $\ge \sqrt {{{\left( {\frac{{x - z}}{2}} \right)}^2} + {{\left[ {\frac{{\left( {x + z} \right)\sqrt 3 }}{2}} \right]}^2}} = \sqrt {{x^2} + xz + {z^2}}$ .

Proof 2. Let $\triangle VBA$ , $\triangle VBC$ where $VB$ separates $A$ , $C$ and $\left\{\begin{array}{ccc} VA=x & VB=y & VC=z\\\\ m\left(\widehat{AVB}\right)=60^{\circ} & m\left(\widehat{CVB}\right)=60^{\circ} & m\left(\widehat{AVC}\right)=120^{\circ}\end{array}\right\|$ .

Observe that $\left\{\begin{array}{c} AB^2=x^2-xy+y^2\\\\ CB^2=z^2-zy+y^2\\\\ AC^2=x^2+xz+z^2\end{array}\right\|$ and our inequality is equivalently with $AB+CB\ge AC$ , what is truly. We have equality iff $B\in (AC)\ \iff$

in $\triangle AVC$ the edge $[VB]$ is the $V$ -bisector in $\triangle AVC\iff$ $\frac {BA}{BC}=\frac {VA}{VC}\iff$ $\frac {\sqrt{y^2-yx+x^2}}{y^2-yz+z^2}=\frac xz\iff$ $z^2\left(y^2-yx+x^2\right)=x^2\left(y^2-yz+z^2\right)\iff$

$y^2\left(x^2-z^2\right)=xyz(x-z)\iff$ $x=z\ \vee\ y(x+z)=xz\iff$ $x=z\ \vee\ \frac 1y=\frac 1x+\frac 1z\iff$ $\boxed{\frac 1y=\frac 1x+\frac 1z}$ . Indeed, if $x=z$ , then the inequality

becomes $2\sqrt {x^2-xy+y^2}\ge x\sqrt 3$ $\iff$ $x^2-4xy+4y^2\ge 0\iff$ $(x-2y)^2\ge 0$ . Obtain equality $\iff$ $\frac x2=\frac y1=\frac z2$ and in this case $\frac 1y=\frac 1x+\frac 1z$ .

Proof 3. $\sqrt{x^2-xy+y^2}+\sqrt{y^2-yz+z^2}\ge \sqrt{x^2+xz+z^2}\iff$

$x^2-xy+y^2+y^2-yz+z^2+2\sqrt{(x^2-xy+y^2)(y^2-yz+z^2)}\geq x^2+xz+z^2\implies$

$2\sqrt{(x^2-xy+y^2)(y^2-yz+z^2)}\geq xy+yz+xz-2y^2\implies$

$4(x^2-xy+y^2)(y^2-yz+z^2)\geq (xy+yz+az-2y^2)^2\implies$ $3(xy-xz+yz)^2\geq 0$ .

An easy extension. $\left\|\begin{array}{c} \{x,y,z,m\}\subset\mathbb R\\\\ 0\ \le\ m\ \le\ 2\end{array}\right\|\ \implies\ \sqrt{x^2-mxy+y^2}+$ $\sqrt{y^2-myz+z^2}\ge$ $\sqrt{x^2+\left(2-m^2\right)xz+z^2}$ with equality iff $\frac 1x+\frac 1z=\frac my$ .

Proof. Let $\triangle VBA$ , $\triangle VBC$ where $VB$ separates $A$ , $C$ and $\left\{\begin{array}{ccc} VA=x\ \ \wedge\ \ VB=y\ \ \wedge\ \ VC=z\\\\ m\left(\widehat{AVB}\right)=m\left(\widehat{CVB}\right)=\phi\end{array}\right\|$ , where $\phi\in\left[0,\frac {\pi}{2}\right]$ . Denote $m=2\cos\phi\in [0,2]$

Observe that $2\cos 2\phi =2\left(2\cos^2\phi -1\right)=m^2-2$ and $\left\{\begin{array}{c} AB^2=x^2-mxy+y^2\\\\ CB^2=z^2-mzy+y^2\\\\ AC^2=x^2+\left(2-m^2\right)xz+z^2\end{array}\right\|$ and our inequality is equivalently with $AB+CB\ge AC$ ,

what is truly. We have equality $\iff B\in (AC)\ \iff$ in $\triangle AVC$ the edge $[VB]$ is the $V$ -bisector in $\triangle AVC\iff$

$\frac {BA}{BC}=\frac {VA}{VC}\iff$ $\frac {\sqrt{y^2-myx+x^2}}{y^2-myz+z^2}=\frac xz\iff$ $z^2\left(y^2-myx+x^2\right)=x^2\left(y^2-myz+z^2\right)\iff$ $y^2\left(x^2-z^2\right)=mxyz(x-z)\iff$

$x=z\ \vee\ y(x+z)=mxz\iff$ $x=z\ \vee\ \frac my=\frac 1x+\frac 1z\iff$ $\boxed{\frac my=\frac 1x+\frac 1z}$ .

PP8. Let $a,b,c$ be positive real numbers. Prove that $a^3 +b^3 +c^3\ \geq\ a^2b +b^2c +c^2a$ .

Proof. $\left\{\begin{array}{ccc} a^3+a^3+b^3 & \ge & 3\cdot a^2b\\\\ b^3+b^3+c^3 & \ge & 3\cdot b^2c\\\\ c^3+c^3+a^3 & \ge & 3\cdot c^2a\end{array}\right\|\ \bigoplus\ \implies\ a^3+b^3+c^3\ \ge\ a^2b+b^2c+c^2a$ .

An easy extension. Let $a,b,c$ be positive real numbers. Prove that $a^{n+1}+b^{n+1} +c^{n+1} \geq\ a^nb +b^nc +c^na$ for any $n\in\mathbb N$ .

Proof. $\left\{\begin{array}{ccc} \underbrace{a^{n+1}+a^{n+1}+\ \ldots\ +a^{n+1}}_n+b^{n+1} & \ge & (n+1)\cdot a^nb\\\\ \underbrace{b^{n+1}+b^{n+1}+\ \ldots\ +b^{n+1}}_n+c^{n+1} & \ge & (n+1)\cdot b^nc\\\\ \underbrace{c^{n+1}+c^{n+1}+\ \ldots\ +c^{n+1}}_n+a^{n+1} & \ge & (n+1)\cdot c^na\end{array}\right\|\ \bigoplus$ $\implies\ a^{n+1}+b^{n+1} +c^{n+1} \geq\ a^nb +b^nc +c^na$ for any $n\in\mathbb N$ .

PP9. Let $a,b,c$ be three positive numbers . Prove that $\left\{\begin{array}{cccc} 8(a+b+c)(ab+bc+ca) & \le & 9(b+c)(c+a)(a+b) & (1)\\\\ 2\cdot \sqrt{bc+ca+ab} & \leq & \sqrt{3}\cdot \sqrt[3]{(b+c)(c+a)(a+b)} & (2)\\\\ 8\sqrt {abc(a+b+c)(ab+bc+ca)} & \le & 3(b+c)(c+a)(a+b) & (3)\end{array}\right\|$ .

Proof. Denote $\left\{\begin{array}{ccc} a+b+c & = & s_1\\\ ab+bc+ca & = & s_2\\\ abc & = & s_3\end{array}\right\|$ , i.e. $(t-a)(t-b)(t-c)=t^3-s_1\cdot t^2+s_2\cdot t-s_3=0\begin{array}{cc} \nearrow & a\\\ \rightarrow & b\\\ \searrow & c\end{array}$ .

Observe that $s_1s_2=(a+b+c)(ab+bc+ca)\ge$ $3\sqrt [3]{abc}\cdot 3\sqrt [3]{a^2b^2c^2}=$ $9abc=9s_3\implies \boxed{s_1s_2\ge 9s_3}\ (*)$ .

Therefore, $(b+c)(c+a)(a+b)=(s_1-a)(s_1-b)(s_1-c)=$ $s_1^3-s_1^2\cdot s_1+s_1s_2-s_3=s_1s_2-s_3\implies$

$9(b+c)(c+a)(a+b)=9s_1s_2-9s_3\stackrel{(*)}{\ge}9s_1s_2-s_1s_2=8s_1s_2\implies$ $\boxed{9(b+c)(c+a)(a+b)\ge 8(a+b+c)(ab+bc+ca)}\ (1)$ .

In conclusion, $\left\{\begin{array}{ccc} 81(b+c)^2(c+a)^2(a+b)^2 & \ge & 64(a+b+c)^2(ab+bc+ca)^2\\\\ (a+b+c)^2 & \ge & 3(ab+bc+ca)\end{array}\right\| \bigodot\ \implies$

$27(b+c)^2(c+a)^2(a+b)^2\ge 64(ab+bc+ca)^3\iff$ $\boxed{\sqrt{3}\cdot \sqrt[3]{(b+c)(c+a)(a+b)}\ge 2\cdot \sqrt{bc+ca+ab}}\ (2)$ .

$\left\{\begin{array}{ccc} 81(b+c)^2(c+a)^2(a+b)^2 & \ge & 64(a+b+c)^2(ab+bc+ca)^2\\\\ (a+b+c)^2 & \ge & 3(ab+bc+ca)\\\\ (ab+bc+ca)^2 & \ge & 3abc(a+b+c)\end{array}\right\| \bigodot\ \implies$

$9(b+c)^2(c+a)^2(a+b)^2\ge 64abc (a+b+c)(ab+bc+ca)\iff$ $\boxed{3(b+c)(c+a)(a+b)\ge 8\cdot \sqrt{abc(a+b+c)bc+ca+ab)}}\ (3)$ .

PP10. Prove that for any $\{a,b,c\}\subset\mathbb R^*_+$ there is the inequality $2\cdot \sum a^6+16\cdot \sum b^3c^3\ge 9\cdot\sum b^2c^2\left(b^2+c^2\right)$ .

Proof. I"ll show that $b^6+c^6+16b^3c^3\ge 9b^2c^2\left(b^2+c^2\right)\ (*)$ . Indeed, using the substitution $\frac bc=t$ the relation $(*)$ becomes $t^6+1+16t^3-9t^2\left(t^2+1\right)\ge 0$ . Using the substitution

$t+\frac 1t=y$ obtain that $t^3+\frac {1}{t^3}+16-9\left(t+\frac 1t\right)\ge 0\iff$ $y^3-3y+16-9y\ge 0\iff$ $y^3-12y+16\ge 0\iff$ $(y-2)^2(y+4)\ge 0$ , what is truly. I have equality

if and only if $y=2\iff t=1\iff b=c$ . In conclusion, $\left\{\begin{array}{c} b^6+c^6+16b^3c^3\ge 9b^2c^2\left(b^2+c^2\right)\\\\ c^6+a^6+16c^3a^3\ge 9c^2a^2\left(c^2+a^2\right)\\\\ a^6+b^6+16a^3b^3\ge 9a^2b^2\left(a^2+b^2\right)\end{array}\right\|\ \bigoplus\ \implies 2\cdot \sum a^6+$ $16\cdot \sum b^3c^3\ge 9\cdot\sum b^2c^2\left(b^2+c^2\right)$ .

PP11. Find the maximum value of $\left|\sqrt{x^2+4x+5}-\sqrt{x^2+2x+5}\right|^4$ .

Proof 1 $.\ \left|\sqrt{x^2+4x+5}-\sqrt{x^2+2x+5}\right|=\left|\sqrt{(x+2)^2+(0-1)^2}-\sqrt{(x+1)^2+(0-2)^2}\right|$ . Let a mobile

$P(x,0)\in\mathrm{Ox}$ and fixed $A(-2,1)$ , $B(-1,2)$ . Now $\left|\sqrt{x^2+4x+5}-\sqrt{x^2+2x+5}\right|=|PA-PB|\le AB=\sqrt 2$ .

Have the equality for $P\in AB\cap\mathrm{Ox}$ $\iff P(-3,0)$ . Hence $\max \left|\sqrt{x^2+4x+5}-\sqrt{x^2+2x+5}\right|^4= 4 \text{ at }x=-3$ .

PP12. Prove that $\{x,y,z\}\subset\mathbb R\implies 8\cdot \large{(x^3+y^3+z^3)^2 \ge 9\cdot (x^2+yz)(y^2+zx)(z^2+xy)}$ .

Proof. $27\cdot (x^2+yz)(y^2+zx)(z^2+xy) \le (x^2+y^2+z^2+xy+yz+zx)^3 \le$ $8\cdot (x^2+y^2+z^2)^3 \le 24\cdot (x^3+y^3+z^3)^2$ .

Remark. I used the remarkable power main inequality $\sqrt[2]{\frac {x^2+y^2+z^2}{3}}\le \sqrt [3]{\frac {x^3+y^3+z^3}{3}}$ .

PP13. Let $a>0$ , $b>0$ and $a+b=1$ . Prove that $2<\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)\le\frac{9}{4}$ .

Proof. $\{a,b\}\subset \mathbb R^*_+$ and $a+b=1\implies$ $\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)=$ $\frac {\left(1-a^2\right)\left(1-b^2\right)}{ab}=$ $\frac {(1+a)(1-a)(1+b)(1-b)}{ab}=$

$\frac {(1+a)b(1+b)a}{ab}=$ $(1+a)(1+b)=1+(a+b)+ab=2+ab$ . In conclusion, $2<2+ab\le 2+\left(\frac {a+b}{2}\right)^2=\frac{9}{4}$ .

PP14. Prove that for $a_k>0\ (k=1,\ 2,\ \cdots n)$ , $\sum_{k=1}^n a_k^2+\frac{n(n+1)(2n+1)}{6}\geq 2\sum_{k=1}^n ka_k$ .

Proof. $\sum_{k=1}^n\left(a_k-k\right)^2\ge 0\iff$ $\sum_{k=1}^na_k^2+\sum_{k=1}^nk^2\ge 2\sum_{k=1}^nka_k\iff$ $\sum_{k=1}^n a_k^2+\frac{n(n+1)(2n+1)}{6}\geq 2\sum_{k=1}^n ka_k$ .

PP15. Let $\in\left(0,\frac{\pi}{2}\right)$ . Prove that $\frac{3\sqrt{3}}{\sin x}+\frac{1}{\cos x}\ge 8$ .

Proof 1. Observe that $2\left(\frac {3\sqrt 3}{\sin x}+\frac 1{\cos x}\right)\ge \left(\sqrt 3\sin x+\cos x\right)\left(\frac {3\sqrt 3}{\sin x}+\frac 1{\cos x}\right)=$

$\left(\frac {\sqrt 3}{\sin x}+\frac {\sqrt 3}{\sin x}+\frac {\sqrt 3}{\sin x}+\frac 1{\cos x}\right)\left(\frac {\sin x}{\sqrt 3}+\frac {\sin x}{\sqrt 3}+\frac {\sin x}{\sqrt 3}+\cos x\right)\ge 4^2=16$ $\implies$ $\frac{3\sqrt{3}}{\sin x}+\frac{1}{\cos x}\ge 8$ .

Proof 2 (ugly). Observe that $\frac {3\sqrt 3}{\sin x}+\frac 1{\cos x}\ge 8\iff$ $\left(\sin^2x+\cos^2x\right)\left(3\sqrt 3\cos x+\sin x\right)^2\ge$ $64\sin^2x\cos^2x\ \stackrel{\tan x=t}{\iff}$ $\left(t^2+1\right)$ $\left(t+3\sqrt 3\right)^2\ge 64t^2\iff$

$t^4+6\sqrt 3t^3-36t^2+6\sqrt 3t+27\ge 0\iff$ $\left(t-\sqrt 3\right)^2\left(t^2+8\sqrt 3t+9\right)\ge 0$ , what is truly. We have the equality if and only if $t=\sqrt 3$ , i.e. $\tan x=\sqrt 3\iff x=\frac {\pi}{3}$ .

Proof 3. $\frac{3\sqrt{3}}{\sin x}+\frac{1}{\cos x}=$ $\frac{\sqrt{3}}{\sin x}+\frac{\sqrt{3}}{\sin x}+\frac{\sqrt{3}}{\sin x}+\frac{1}{\cos x}\ge$ $4\cdot\sqrt[4]{\frac {3\sqrt3\left(\sin^2x+\cos^2x\right)^2}{\sin^3x\cos x}}$ $\stackrel{(\tan x=t)}{=}$ $4\cdot\sqrt[4]{\frac {3\sqrt3\left(t^2+1\right)^2}{t^3}}$ . Therefore, $\frac{3\sqrt{3}}{\sin x}+\frac{1}{\cos x}\ge 8\iff$

$3\sqrt 3\left(t^2+1\right)^2\ge 16t^3\iff$ $3\sqrt 3t^4-16t^3+6\sqrt 3t^2+3\sqrt 3\ge 0\iff$ $\left(t-\sqrt 3\right)^2\left(3\sqrt 3t^2+2t+\sqrt 3\right)\ge 0$ , it is true. Have equality iff $\tan x=t=\sqrt 3$ , i.e. $x=60^{\circ}$ .

An easy extension (sqing). Prove that for any $\{a,b,u,v\}\in \mathbb R^*_+$ there is the inequality $\boxed{\frac {a}{u}+\frac b{v}\ge \frac {(a+b)^2}{\sqrt {\left(a^2+b^2\right)\left(u^2+v^2\right)}}}$ .

Proof. $\frac au+\frac bv=\frac {a^2}{au}+$ $\frac {b^2}{bv}\ \stackrel{C.B.S}{\ge}\ \frac {(a+b)^2}{au+bv}\ \stackrel{C.B.S}{\ge}\ \frac {(a+b)^2}{\sqrt {\left(a^2+b^2\right)\left(u^2+v^2\right)}}$ . Particular case. If $\{a,b,u,v\}\in \mathbb R^*_+$ and $x\in\left(0,\frac {\pi}{2}\right)$ , then $\frac {a^2}{u\sin x}+\frac {b^2}{v\cos x}\ge \frac{(a+b)^2}{\sqrt {u^2+v^2}}$ .

PP16. Prove that the implication: $x\in\left(0,\pi\right)\ \implies\ \sin x(1-\cos x)\le\frac {3\sqrt 3}{4}$ .

Proof 0 (with derivatives). Denote $f(x)=\sin x(1-\cos x)\ ,\ x\in (0,\pi )$ . Prove easily that $f'(x)=-2\cos^2x+\cos x+1$ $\iff$ $\boxed{f'(x)=(1-\cos x)(2\cos x+1)}$

and $f'(x)=0\iff x=\frac {2\pi}{3}\in (0,\pi)$ . In conclusion $,\ \left\|\begin{array}{c} x\\\\ f'\\\\ f\end{array}\right|$ $\left|\begin{array}{ccc} & \frac {2\pi}{3} & \\\\ + & 0 & -\\\\ \nearrow & \frac {3\sqrt 3}{4} & \searrow\end{array}\right\|\implies$ $\boxed{\max_{0<x<\pi}f(x)=f\left(\frac {2\pi}{3}\right)=\frac {3\sqrt 3}{4}}$ , i.e. $x\in\left(0,\pi\right)\implies\sin x(1-\cos x)\le\frac {3\sqrt 3}{4}$ .

Proof 1. $\sin x(1-\cos x)$ is maximum $\iff$ $\sin^2x(1-\cos x)^2$ is maximum $\iff$ $(1+\cos x)(1-\cos x)^3$ is maximum $\iff$ $(1+\cos x)\cdot \frac {1-\cos x}{3}\cdot\frac {1-\cos x}{3}\cdot\frac {1-\cos x}{3}$

is maximum . Since $(1+\cos x)+\frac {1-\cos x}{3}+\frac {1-\cos x}{3}+\frac {1-\cos x}{3}=2$ (constant) obtain that $\sin x(1-\cos x)$ is maximum $\iff$ $1+\cos x=\frac {1-\cos x}{3}=\frac 24\iff$

$\cos x=-\frac 12\iff$ $x=\frac {2\pi}{3}$ . With othet words, $\sin x(1-\cos x)\le \sin\frac {2\pi}{3}\left(1-\cos \frac {2\pi}{3}\right)\iff$ $\sin x(1-\cos x)\le\frac {3\sqrt 3}{4}$ .

Proof 2. $\sin x(1-\cos x)=4\sin^3\frac x2\cos\frac x2=$ $\frac {4\sin^3\frac x2\cos\frac x2}{\left(\sin^2\frac x2+\cos^2\frac x2\right)^2}=$ $\frac {4t^3}{\left(t^2+1\right)^2}$ , where $t=\tan\frac x2\in(0,\infty)$ . Thus, $\sin x(1-\cos x)\le\frac {3\sqrt 3}{4}\iff$

$3\sqrt 3\left(t^2+1\right)^2-16t^3\ge 0\iff$ $\left(t-\sqrt 3\right)^2\left(3\sqrt 3t^2+2t+\sqrt 3\right)\ge 0$ , what is truly. Have equality if and only if $t=\sqrt 3\iff$ $\tan\frac x2=\sqrt 3\iff$ $x=\frac {2\pi}{3}$ .

An easy extension. Prove that the implication: $\{m,n\}\subset\mathbb N^*\ ,\ x\in\left(0,\pi\right)\ \implies\ \sin^m x(1-\cos x)^n\le\frac {(m+2n)^n\sqrt {m(m+2n)}}{(m+n)^{n+1}}$ .

This post has been edited 157 times. Last edited by Virgil Nicula, Nov 21, 2015, 5:43 PM

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327. Some geometrical/algebraic extremum problems.

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315. Indian Team Selection Test 2010 ST1 P1 and extension.

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312. Very nice (famous trigonometrical problems) !

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309. Beautiful problem ! IRAN Selection Test for IMO, 2004.

308. Some nice and easy properties in a triangle.

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307. Very nice proofs !

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289. Nice identities in an algebra/ geometry/trigonometry.

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265. Really nice problem !

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258. Some difficult and nice problems with complex numbers.

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256. An extremum problems with complex numbers.

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255. A line which is parallelly with OH.

254. An interesting C. Mateescu's geometrical inequality.

253. An old algebraic inequality.

252. OI=f(A,R) in certain conditions.

251. Problems of Analytical Geometry.

250. An application of the Euler's relation.

249. An usual and nice metrical relations in quadrilateral.

248. Pagina instructiva.

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246. R1-R2=OI (nice).

245. An inequality in an acute triangle.

244. Some properties of cyclic orthodiagonal quadrilaterals.

243. Nice ! (a+b) is constant.

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241. Areas in a triangle.

240. Some interesting geometrical inequalities.

239. Algebraic marathon.

238. Some metrical problems.

237. Some problems with ... problems.

236. A locus with the metrical relation.

235. Proofs of some geometry problems with complex numbers.

234. Some simple and nice problems from contests.

233. An interesting identity and an easy & nice inequality.

February 2011

232. Some geometrical inequalities (own).

Pagina libera

231. A conditioned minimum value.

230. Some usual synthetical properties in a triangle.

229. Concurs online MATEFBC, ed. 5 -subiect 2.

228. OJM - Romania, 2010 problem 3.

227. IMO ShortList 2003 (problem 5).

226. Some inequalities from Calculus (Real Analysis).

225. An interesting problems from Kunny.

224. Some nice and easy inequalities in a triangle.

223. An inequality with complex numbers-RMO shortlist 2010.

222. An extension of probl. 2 from IMO 2009.

221. Some nice and simple "slicing" problems.

220. Some nice geometry problems from contest.

January 2011

219. A very nice geometrical inequality.

218. Some problems from the Suiss IMO Selection Team 2006.

217. A nice and interesting Murray S. Klamkin's problem.

216. Nice ! Find the length of hypotenuse (Belarus - 2010)

215. Limit of the unique root for polynominal equations.

214. A range of some functions.

213. Another classical "slicing" problems.

212. A simple applications of the Casey's theorem.

211. Pretty hard problem !

210. Tangential circle of the triangle ABC.

209. Nice problem, nice proof !

208. An interesting geometrical inequality.

207. Some interesting problems of Toshio Seimiya, Japan.

206. Four problems with extremum in a quadrilateral.

205. A "slicing" problem with 11 methods.

204. Saraghin 2010 (three geometry problems).

203. An application of the Pascal's permutation theorem.

December 2010

202. Concurrent lines in a right triangle.

201. Applications of Desarques' and Pappus' theorems.

200. Some nice applications of the inversion.

199. Some properties of symmedian. Two extensions.

198. Generalized Carnot's lemma.

197. Triangles outside of a triangle and concurrence.

195. Semicircle. Nice application of harmonical division.

194. The Lalescu's sequence. Properties.

193. An inequality with n! .

192. An easy and nice problem of Valentin Vornicu.

191. Six nice problems with the incircle of a triangle.

190. Very nice ! Coculescu's Contest seniors 2010.

189. Pascal's theorem and some nice and easy applications.

188. Some problems from vectorial geometry.

187. Distancies.

186. Without the l'Hospital's rules.

185. The Durrande's relation.

184. An interesting geometry problem from RMO 2010.

183. A remarkable characterization of equilateral triangle.

182. Another two nice problems with complex numbers (own).

181. Two special inequalities from CRUX (own).

180. Own proposed problem from CRUX (with complex numbers).

November 2010

179. The range of |z| , |z^2+a|=|bz+c|. Particular cases.

178. Some nice and simple geometry problems.

177. A nice proof of one identity in a triangle.

176. Another nice problems for middle school.

175. A Geometric Proof (without trig.) of Heron's Formula.

174. Some interesting properties of modulus (middle school).

173. An interesting trigonometric equations.

172. The Zelinsky's problem.

171. Another "slicing" proposed problems (middle school).

October 2010

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

169. Some metrical problems in a circle (III).

168. Some properties of the Lemoine's point.

167. Nice problems - OIM, 1995 and ... Toshio Seimyia.

166. Some constant geometrical expressions.

165. Some metrical problems (5) in a circle (II).

164. Some metrical problems (5) in a circle.

163. Applications of Ptolemy's inequality. Ex. - China,1998.

162. Generalization of proposed problem from Mongolia, 2000.

161. A difficult inequality (barycentrical coordinates)

160. A very nice "slicing" problem.

159. Multiple derivatives in a point.

158. Very nice and difficult limits.

157. Steinbart's theorem. Applications.

156. Some problems with perpendicularities.

155. A constant ratio in a triangle ABC.

154. Difficult inequality (without derivatives eventually).

153. Geometrical minimum/maximum.

152. Concurrence, cyclically, metrics, symmedian a.s.o.

151. Miscellaneous problems for the admission at math.

150. Some (5) problems with fixed points.

149. Some (8) problems with collinearity or concurrence.

148. A special class of systems.

147. A problem with two polynominals.

146. Synthetic, metric, trigonometric and analitycal proofs.

145. Spain MO P5 - Point on median.

144. A conditioned concurrency in a triangle.

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

142. Inscribed/circumscribed quadrilateral w.r.t. ABC.

141. N is interior <==> 1/3<a/b<3 (CRUX Mathematicorum).

140. A synthetical property <==> linear/angled relation.

139. A caracterization of a right angle in a triangle.

138. Generalized "slicing" problems (own).

137. Stewart's relation.

136. A geometrical interpretation of a system (own).

135. Three strong geometrical inequalities.

September 2010

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

133. Some problems with projections/perpendiculars (1).

132. Asupra unei inegalitati intr-un triunghi.

131. A stronger Form of the Steiner-Lehmus Theorem (own).

130. A triangle and a fixed point.

129. BX=CY and three properties.

128. Value of IG and two inequalities in neighbourhood of 0.

127. Pagina educativa.

126. Routh's theorem.

125. I like very much this Darij Grinberg's message.

124. A nice and easy problem with a right-angled triangle.

123. My extension and its proof.

122. An identity for an equilateral triangle. Extension.

121. Viete's relations with cos A , cos B , cos C.

120. Two equal neighbouring angles/sides in a quadrilateral.

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

orzzzzzzzzz
by mathMagicOPS, Jan 9, 2025, 3:40 AM
this css is sus
by ihatemath123, Aug 14, 2024, 1:53 AM
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
by GoogleNebula, Apr 14, 2021, 5:25 AM
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

285. Nice and easy inequalities. For example, IMAC - 2009.

by Virgil Nicula, Jun 13, 2011, 1:42 PM

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