364. Some interesting inequalities I

by Virgil Nicula, Jan 16, 2013, 9:21 PM

PP1. Let $a$ , $b$ and $c$ be the sidelengths of $\triangle ABC$ with the inradius $r$ and $2s=a+b+c$ . Prove that $\boxed{\cos\frac A2\le \frac{\sqrt{bc}}{4r}}$ and $\boxed{\cos\frac {B-C}{2}\ge \sqrt {\frac {2r}{R}}}$ .

Proof 1. $\left\{\begin{array}{c} b^2\ge b^2-(a-c)^2=4(s-a)(s-c)\\\\ c^2\ge c^2-(a-b)^2=4(s-a)(s-b)\end{array}\right|\bigodot\implies$ $b^2c^2\ge 16(s-a)^2(s-b)(s-c)=16(s-a)sr^2\implies$ $bc\ge 4r\sqrt {s(s-a)}\implies$ $\sqrt {bc}\ge 4r\sqrt{\frac {s(s-a)}{bc}}\implies$

$\sqrt {bc}\ge 4r\cos\frac A2\implies$ $\boxed{\cos\frac A2\le \frac{\sqrt{bc}}{4r}}$ . Observe that $\cos\frac {B-C}{2}\ge \sqrt {\frac {2r}R}$ and $\sin\frac {B+C}{2}=\cos\frac A2\implies$ $2\sin\frac {B+C}{2}\cos\frac {B-C}{2}\ge 2\cos\frac A2\sqrt{\frac {2r}R}\iff$

$\sin B+\sin C\ge 2\sqrt{\frac {2rs(s-a)}{Rbc}}\iff$ $b+c\ge\sqrt{\frac {32Rrs(s-a)}{bc}}\iff$ $(b+c)^2\ge 8a(s-a)\iff$ $[a+(b+c-a)]^2\ge 4a(b+c-a)\iff$ $[a-(b+c-a)]^2\ge 0$ ,

what is truly. Have equality iff $\boxed{b+c=2a\iff IO\perp IA}$ .

Proof 2.. Let $P\in (AC)$ so that $IP\perp AC$ , the midpoint $M$ of $[BC]$ , the incircle $w=C(I,r)$ , the circumcircle $C(O,R)$ and the diameter $[NS]$ , where $M\in NS$ and $BC$ separates

$A\ ,\ S$ . So $SC=IS$ and $\triangle AIP\sim\triangle NSC\implies$ $\frac {IA}{SN}=\frac {IP}{SC}\implies$ $\frac {IA}{2R}=\frac r{SC}\implies$ $\boxed{-p_w(I)\equiv IA\cdot IS=2Rr}$ . Thus, $2R\cos\frac {B-C}{2}=AS=AI+IS\ge$

$2\sqrt {IA\cdot IS}=2\sqrt {2Rr}\implies$ $\cos\frac {B-C}{2}\ge\sqrt{\frac{2r}{R}}$ .

Proof 3.. Let $D\in AI\cap BC$ . Thus, $\triangle ABS\sim\triangle ADC\iff$ $\frac {AS}{AB}=\frac {AC}{AD}\iff$ $\boxed{AD\cdot AS=bc}\ (*)$ . Therefore, $\frac {IA}{b+c}=\frac {ID}{a}=\frac {AD}{2s}$ and

from the Ptolemy theorem obtain that $a\cdot AS=(b+c)\cdot IS\implies$ $IA\cdot IS=IA\cdot\frac {a\cdot AS}{b+c}=$ $AS\cdot \frac {a}{2s}$ $\cdot AD\ \stackrel{(*)}{=}\ \frac {abc}{2s}$ $\implies$ $IA\cdot IS=2Rr$ .

Proof 4, Observe that $SI=SB=SC=2R\sin\frac A2$ and $AI^2=\frac {bc(s-a)}{s}$ . Thus, $IA\cdot IS=$ $\sqrt{\frac {bc(s-a)}{s}}\cdot 2R\sqrt {\frac {(s-b)(s-c)}{bc}}=$

$2R\cdot\sqrt{\frac {(s-a)(s-b)(s-c)}{s}}=2Rr\implies$ $IA\cdot IS=2Rr$ .

PP2. Let $a$ , $b$ and $c$ be the sidelengths of an acute $\triangle ABC$ . Prove that $\boxed{4a\cos A\le\sqrt{\frac{bc}{\cos B\cos C}}}\ (1)$ , i.e. $\boxed{2\sin 2A\le\sqrt {\tan B\tan C}}$ .

Proof 1. Let $XX$ be the tangent to the circumcircle $w$ of $\triangle ABC$ in $X\in w$ . Let $\triangle MNP$ , where $\left\{\begin{array}{c} M\in BB\cap CC\ ;\ M=\pi -2A\ ;\ NP=m\\\\ N\in CC\cap AA\ ;\ N=\pi-2B\ :\ PM=n\\\\ P\in AA\cap BB\ ;\ P=\pi -2C\ ;\ MN=p\end{array}\right|$ and $\frac {m}{a\cos A}=\frac {n}{b\cos B}=$

$\frac {x+y+z}{2}\ge \frac 32\sqrt[3]{xyz}=\frac 32$ . $\frac {p}{c\cos C}=\frac {1}{2\cos A\cos B\cos C}$ . Thus, $(*)$ becomes $\cos\frac M2\le\frac {\sqrt {np}}{4R}\iff$ $4R\sin A\le\frac 1{2\cos A}\sqrt {\frac {bc}{\cos B\cos C}}\iff$ $4a\cos A\le\sqrt{\frac{bc}{\cos B\cos C}}$ .

Proof 2. Observe that $b^2+c^2-a^2>0$ , $a^2+c^2-b^2>0$ , $a^2+b^2-c^2>0$ and $\left\{\begin{array}{c} (b^2+c^2-a^2)\cdot (a^2+c^2-b^2)\le \left[\frac {(b^2+c^2-a^2)+(a^2+c^2-b^2)}{2}\right]^2=c^4\\\\ (b^2+c^2-a^2)\cdot (a^2+b^2-c^2)\le \left[\frac {(b^2+c^2-a^2)+(a^2+b^2-c^2)}{2}\right]^2=b^4\end{array}\right|\ \bigodot$ $\implies$

$\left(b^2+c^2-a^2\right)^2\left(a^2+c^2-b^2\right)\left(a^2+b^2-c^2\right)\le b^4c^4\iff$ $(2bc\cos A)^2\cdot 2ac\cos B\cdot 2ab\cos C\le b^4c^4\iff$ $16a^2\cos^2A\cos B\cos C\le bc\iff$

$16\sin^2A\cos^2A\cos B\cos C\le \sin B\sin C\iff$ $4\sin^22A\le\tan B\tan C\iff$ $2\sin 2A\le \sqrt {\tan B\tan C}$ .

Remark. $(1)\iff$ $2\sin 2A\le\sqrt {\tan B\tan C}$ . If denote $\left\{\begin{array}{c} \tan A=x\\\\ \tan B=y\\\\ \tan C=z\end{array}\right|$ , then $\boxed{\left|\begin{array}{c} \{x,y,z\}\subset\mathbb R^*_+\\\\ x+y+z=xyz\end{array}\right\|\implies \frac {4x}{x^2+1}\le\sqrt {yz}}$ .

Proof. By AM-GM $xyz=x+y+z\ge 3\sqrt[3]{xyz} \implies xyz \ge 3\sqrt3$ . Hence $\frac{4x}{1+x^2}=$ $\frac{4x}{1+\frac{x^2}3+\frac{x^2}3+\frac{x^2}3}\le$ $\frac{4x}{4\sqrt[4]{\frac{x^6}{27}}}=$ $\frac{\sqrt[4]{27}}{\sqrt{x}} =$ $\frac{\sqrt[4]{27}\cdot\sqrt{yz}}{\sqrt{xyz}}\le$ $\frac{\sqrt[4]{27}\cdot\sqrt{yz}}{\sqrt[4]{27}} =$ $\sqrt{yz}$ .

PP3. Prove that in any triangle $ABC$ there is the inequality $\boxed{\frac {bc}{a}+\frac {ca}{b}+\frac {ab}{c}\ge a+b+c\ge \frac {4S}{R}}$ .

Proof 1. $\frac {bc}{a}+\frac {ca}{b}+\frac {ab}{c}\ge \frac {4S}{R}\iff$ $R\cdot\sum b^2c^2\ge 4abcS\iff$ $\sum b^2c^2\ge 16S^2\iff$ $\sum b^2c^2\ge 2\sum b^2c^2-\sum a^4\iff$ $\sum a^4\ge \sum b^2c^2$ , what is truly.

Proof 2. $\left\{\begin{array}{c} 4(s-a)(s-b)\le c^2\\\\ 4(s-a)(s-c)\le b^2\end{array}\right|\bigodot\implies$ $16(s-a)sr^2\le b^2c^2\iff$ $4(s-a)rabc\le b^2c^2R\iff$ $\frac {bc}{a}\ge \frac {4r(s-a)}{R}\implies$ $\sum \frac {bc}{a}\ge \frac {4rs}{R}=\frac {4S}{R}$ .

Proof 3. $\frac bc+\frac cb\ge 2\implies$ $\sum \frac {bc}{a}=\frac 12\cdot\sum \left(\frac {ac}{b}+\frac {ab}{c}\right)\ge a+b+c=2s=\frac {2sR}{R}\ge \frac {4sr}{R}=\frac {4S}{R}$ . I used the well-known inequality $R\ge 2r$ .

Remark. $\frac {bc}{a}+\frac {ca}{b}+\frac {ab}{c}\ge \frac {4S}{R}\iff$ $R\cdot\sum b^2c^2\ge 4abcS\iff$ $\sum b^2c^2\ge 16S^2\iff$ $\sum\frac {1}{a^2}\ge \frac {16S^2}{a^2b^2c^2}\iff$ $\boxed{\frac {1}{a^2}+\frac {1}{b^2}+\frac {1}{c^2}\ge \frac {1}{R^2}}$ .

PP4. Let an acute $\triangle ABC$ with circumcircle $w=C(O,R)$ . $AO$ , $BO$ , $CO$ cut again circumcircle

of $\triangle OBC$ , $\triangle AOC$ , $\triangle ABO$ at $P$ , $Q$ , $R$ respectively. Prove that $OP \cdot OQ \cdot OR\ge 8R^3$ .

Proof. Denote the midpoint $D$ of the side $[BC]$ and the diameter $[OA_1]$ of the circumcircle $w_a=C(O_1, R_1)$ of $\triangle OBC$ . Observe that $OBA_1C$ is a right deltoid and

$BC=2R_1\sin \widehat {BOC}\implies$ $2R\sin A=2R_1\sin 2A\implies$ $R_1=\frac {R}{2\cos A}$ . Otherwise, $OA_1\cdot OD=DB^2\implies$ $2R_1\cdot R\cos A=R^2\implies$ $\boxed{R_1=\frac {R}{2\cos A}}$ a.s.o.

Also, $OP=OA_1\cos\widehat{A_1OP}\implies$ $OP=2R_1\cos (B-C)\implies$ $\boxed{OP=R\cdot \frac {\cos (B-C)}{\cos A}}$ a.s.o. In conclusion, $OP\cdot OQ\cdot OR=$ $R^3\cdot\prod\frac {\cos (B-C)}{\cos A}=$

$R^3\prod\frac {\sin 2B+\sin 2C}{\sin 2A}\ge 8$ because $\{x,y,z\}\subset \mathrm R^*_+$ $\implies$ $\frac {(x+y)(y+z)(z+x)}{xyz}\ge 8$ , where $\left\{\begin{array}{c} \sin 2A=x>0\\\\ \sin 2B=y>0\\\\ \sin 2C=z>0\end{array}\right|$ .

PP5. Prove that $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}} \geq 2$ for any positive numbers $\{a,b,c\}$ .

Proof. Prove easily that $\left[a+(b+c)\right]^2\ge 4a(b+c)$ . Thus, $\frac {a}{b+c}\ge \frac {4a^2}{(a+b+c)^2}\implies$ $\sqrt {\frac {a}{b+c}}\ge\frac {2a}{a+b+c}$ a.s.o.

PP6. Let $a>2$ and $b>0$ . Prove that $\frac ab+\frac {2b}{a-2}-\frac 8a\ge 2$ .

Proof. I"ll show that $\boxed{\sqrt {\frac {2a}{a-2}}\ge 1+\frac 4a}\ (*)$ . Indeed, it is equivalently with $2a^3\ge (a-2)(a+4)^2\iff$ $a^3+32\ge 6a^2$ , what is truly because

$a^3+32=\frac {a^3}{2}+\frac {a^3}{2}+32\stackrel{\mathrm{AM\ge GM}}{\ \ge}$ $3\sqrt [3]{\frac {a^3}{2}\cdot\frac {a^3}{2}\cdot 32}=6a^2$ . Hence $\frac ab+\frac {2b}{a-2}\stackrel{\mathrm{AM}\ge \mathrm{GM}}{\ge}2\sqrt {\frac {2a}{a-2}}\stackrel{(*)}{\ge}$ $2\left(1+\frac 4a\right)$ . In conclusion,

$\frac ab+\frac {2b}{a-2}-\frac 8a\ge 2$ . Have the equality if and only if $a=4\ \wedge\ b=2$ . I applied the inequality AM (arithmetical mean) $\ge$ GM (geometrical mean) for $n\in \{2,3\}$ .

Otherwise. $\sqrt{\frac {2a}{a-2}}\ge 1+\frac 4a\iff$ $2a^3\ge (a-2)(a+4)^2\iff$ $a^3-6a^2+32\ge 0\iff$ $(a+2)(a-4)^2\ge 0$ , what is truly.

Otherwise. Can use directly the identity $2a^3=(a-2)(a+4)^2+(a+2)(a-4)^2\implies$ $2a^3\ge (a-2)(a+4)^2$ .

An easy extension. Let $a>c>0$ and $b>0$ . Prove that $\frac ab+\frac {2b}{a-c}-\frac {4c}a\ge 2$ . For $c=2$ obtain the proposed problem PP6.

Proof 1. $2a^3=(a-c)(a+2c)^2+(a+c)(a-2c)^2\implies$ $2a^3\ge (a-c)(a+2c)^2\implies$ $\frac {2a}{a-c}\ge \left(\frac {a+2c}{a}\right)^2\iff$ $\boxed{\sqrt {\frac {2a}{a-c}}\ge\frac {a+2c}{a}}\ (*)$ .

Therefore, $\frac ab+\frac {2b}{a-c}\stackrel{\mathrm{AM\ge GM}}{\ \ge\ }2\sqrt {\frac {2a}{a-c}}\stackrel{(*)}{\ge}$ $2\left(1+\frac {2c}a\right)$ . In conclusion, $\frac ab+\frac {2b}{a-c}-\frac {4c}a\ge 2$ . Have the equality if and only if $a=2c\ \wedge\ b=c$ .

I applied the inequality AM (arithmetical mean) $\ge$ GM (geometrical mean) for $n\in \{2,3\}$ . For $c:=2$ obtain the proposed problem.

Proof 2. $\frac ab+\frac {2b}{a-c}-\frac {4c}a\ge 2\iff$ $\frac a{2b}+\frac {b}{a-c}\ge 1+\frac {2c}a\iff$ $\frac a{2b}+\frac {b}{a-c}\ge 3-2+\frac {2c}a\iff$ $\frac a{2b}+\frac {b}{a-c}\ge 3-\frac {2(a-c)}a\iff$ $\frac a{2b}+\frac {b}{a-c}+\frac {2(a-c)}a\ge 3$ ,

what is truly because $\frac a{2b}+\frac {b}{a-c}+\frac {2(a-c)}a\ge 3\sqrt{\frac a{2b}}\cdot\frac {b}{a-c}\cdot\frac {2(a-c)}a=3$ is an application of the inequality AM (arithmetical mean) $\ge$ GM (geometrical mean) for $n=3$ .

Equivalent enunciation. $\{a,b,c\}\subset \mathbb R^*_+\implies \frac {a+c}{b}+\frac {2b}{a}\ge 2\left(1+\frac {2c}{a+c}\right)$ .

Proof 1. Using the substitutions $\frac ax=\frac by=c$ our inequality becomes $\frac {x+1}{y}+\frac {2y}{x}\ge 2\left(1+\frac 2{x+1}\right)\iff$ $(x+1)\left(x^2+x+2y^2\right)\ge 2xy(x+3)\iff$

$2(x+1)\cdot y^2-2x(x+3)\cdot y+x(x+1)^2\ge 0$ , what is truly because its discriminant is $\Delta '=x^2(x+3)^2-2x(x+1)^3=-x\left(x^3-3x+2\right)=-x(x+2)(x-1)^2\le 0$ .

Proof 2. $\frac {a+c}{b}+\frac {2b}{a}\ge 2\left(1+\frac {2c}{a+c}\right)\iff$ $\frac {a+c}{2b}+\frac {b}{a}\ge 1+\frac {2c}{a+c}\iff$ $\frac {a+c}{2b}+\frac {b}{a}\ge 1+2-\frac {2a}{a+c}\iff$ $\frac {a+c}{2b}+\frac {b}{a}+\frac {2a}{a+c}\ge 3$ ,

what is truly because $\frac {a+c}{2b}+\frac {b}{a}+\frac {2a}{a+c}\ge 3\sqrt[3]{\frac {a+c}{2b}\cdot\frac {b}{a}\cdot\frac {2a}{a+c}}=3$ is an application of the inequality AM (arithmetical mean) $\ge$ GM (geometrical mean) for $n=3$ .

PP7 (Abdullaev). Prove that in any $\triangle ABC$ there is the inequality $a^2+b^2+c^2\ge 4S\sqrt 3+ \frac{1}{2}\left(\sum |b-c|\right)^2$ .

Proof. Suppose w.l.o.g. $a\ge b\ge c$ . Thus, $\sum a^2\ge 4S\sqrt 3+ \frac{1}{2}\left(\sum |b-c|\right)^2\iff$ $a^2+b^2+c^2\ge 4S\sqrt 3+2(a-c)^2\iff$ $b^2+4ac\ge a^2+c^2+4S\sqrt 3\iff$

$2ac(2-\cos B)\ge 4S\sqrt 3$ $\iff$ $2ac(2-\cos B)\ge 2\sqrt 3ac\cdot\sin B$ $\iff$ $2-\cos B\ge \sqrt 3\cdot\sin B\iff$ $1\ge \frac 12\cdot\cos B+\frac {\sqrt 3}{2}\cdot\sin B\iff$ $\cos\left(B-60^{\circ}\right)\le 1$ , what

is truly. The Abdullaev's inequality is stronger than the Finsler-Hadwiger's inequality $a^2+b^2+c^2\ge 4S\sqrt 3$ .The shortest proof of the Finsler-Hadwiger's inequality is following.

Denote $S=a^2+b^2+c^2$ and apply the Chebyshev's inequality: $\left\{\begin{array}{cc} a^2\ <\ b^2\ <\ c^2 & \nearrow\\\\ S-2a^2\ <\ S-2b^2\ <\ S-3c^2 & \searrow\end{array}\right\|\implies$ $\left(\sum a^2\right)^2=$ $\sum a^2\cdot\sum\left(b^2+c^2-a^2\right)\ge$

$3\sum a^2\left(b^2+c^2-a^2\right)=$ $3\cdot\left(2\sum a^2b^2-\sum a^4\right)=48S^2\implies$ $\boxed{a^2+b^2+c^2\ge4S\sqrt3}$ .

PP8 (O.I.M. 1995). Prove that $\{a,b,c\}\subset\mathbb R^*_+\ ,\ abc=1\ \implies\ \frac {1}{a^3(b+c)}+\frac 1{b^3(c+a)}+\frac 1{c^3(a+b)}\ge \frac 32$ .

Proof. With $a=\frac 1x\ ,\ b=\frac 1y\ ,\ c=\frac 1z\implies$ our inequality becomes $\{x,y,z\}\subset\mathbb R^*_+\ ,\ xyz=1\ \implies$

$\sum\frac {x^2}{y+z}\ge \frac 32$ . I"ll use $\mathrm{C.B.S.}\ :\ \sum\frac {x^2}{y+z}\ge$ $\frac {\left(\sum x\right)^2}{\sum (y+z)}=$ $\frac {x+y+z}{2}\ge$ $\frac 32\cdot\sqrt[3]{xyz}=\frac 32$ .

Lemma. Prove that $\boxed{\sum_{k=1}^n\frac {k+1}{2^k}\ =\ 3-\frac{n+3}{2^n}\ <\ 3}\ ,\ (\forall )\ n\in\mathbb N^*$ .

Proof 1. Denote $S_n(x)=\sum _{k=1}^n(k+1)x^k$ . Observe that $(x-1)S_n(x)=(n+1)x^{n+1}-x-\sum_{k=1}^nx^k=$ $(n+1)x^{n+1}-x\left(1+\sum_{k=1}^nx^{k-1}\right)=$

$(n+1)x^{n+1}-x\left(1+\frac {x^n-1}{x-1}\right)\implies$ $S_n(x)=\frac {(n+1)x^{n+1}(x-1)-x(x-1)-x\left(x^n-1\right)}{(x-1)^2}\implies$ $S_n(x)=\frac {(n+1)x^{n+2}-(n+2)x^{n+1}-x^2+2x}{(x-1)^2}$ .

Hence $\sum_{k=1}^n\frac {k+1}{2^k}=$ $S_n\left(\frac 12\right)=$ $\frac {(n+1)-2(n+2)+3\cdot 2^n}{2^n}=$ $\frac {3\cdot 2^n-(n+3)}{2^n}$ $\implies$ $\sum_{k=1}^n\frac {k+1}{2^k}=$ $3-\frac{n+3}{2^n}$ . Otherwise. $S_n(x)=\left(\sum_{k=1}^nx^{k+1}\right)^{\prime}=\left(\frac {x^{n+2}-x^2}{x-1}\right)^{\prime}=$

$\frac {\left[(n+2)x^{n+1}-2x\right](x-1)-\left(x^{n+2}-x^2\right)}{(x-1)^2}{}\implies$ $\boxed{S_n(x)=\frac {(n+1)x^{n+2}-(n+2)x^{n+1}-x^2+2x}{(x-1)^2}}$ .

PP9. Prove that for any positive integer $n\ ,\ a_n=\sqrt { 1 + \sqrt { 2 + \sqrt {3+\sqrt { \cdots \sqrt { n }}}}} < 3$ .

Proof. Denote $b_k=\sqrt{k + \sqrt { (k+1)+ \sqrt {(k+2) +\sqrt{\cdots \sqrt { n }}}}}\ ,\ k\in\overline{1,n}$ .

Observe that $a_n=b_1\ ,\ b_k^2=k+b_{k+1}\ ,\ k\in\overline{1,n-1}$ and $b_n^2=n$ . Therefore, $a_n=b_1\le \frac {1+b_1^2}{2}=$ $\frac {2+b_2}{2}=$ $\frac 22+\frac {b_2}{2}\le$ $\frac 22+\frac 12\cdot\frac {1+b_2^2}{2}=$ $\frac 22+\frac 12\cdot\frac {3+b_3}{2}=$ $\frac 22+\frac 34+\frac {b_3}{4}\le$

$\frac 22+\frac 34+\frac 14\cdot\frac {1+b_3^2}{2}=$ $\frac 22+\frac 34+\frac 14\cdot\frac {4+b_4}{2}=$ $\frac 22+\frac 34+\frac 48+\frac {b_4}{8}\le$ $\cdots\ \le\frac 22+\frac 34+\frac 48+\ \cdots\ \frac {n}{2^{n-1}}+\frac {b_n}{2^{n-1}}\le$ $\frac 22+\frac 34+\frac 48+\ \cdots\ +\frac {n}{2^{n-1}}+\frac 1{2^{n-1}}\cdot \frac {1+b_n^2}{2}=$

$\frac 22+\frac 34+\frac 48+\ \cdots\ +\frac {n}{2^{n-1}}+\frac {n+1}{2^n}=$ $\sum_{k=1}^n\frac {k+1}{2^k}\stackrel{(\mathrm{lemma})}{=}3-\frac{(n+3)}{2^n}<3$ .

An easy extension. Prove that for any $m>0$ and positive integer $n\ ,\ a_n(m)=\sqrt { m + \sqrt { m+1 + \sqrt {m+2+\sqrt { \cdots \sqrt { m+n-1 }}}}} < m+2$ .

PP10. $\{\ a\ ,\ b\ \}\ \subset\ [\ 0\ ,\ 1\ ]\ \implies\ 1 - \ \frac {a + b}{2}\ + \ \frac {ab}{3}\ \ge\ \frac {1}{1 + a + b}$

Proof. Denote $\left\|\begin{array}{c} a + b = S \\ \\ ab = P\end{array}\right\|$ . The proposed inequality becomes : $\frac {1}{S + 1} + \frac S2\le\frac P3 + 1$ $\Longleftrightarrow$

$3(S^2 + S + 2)\le 2(P + 3)(S + 1)$ $\Longleftrightarrow$ $\boxed {\ 3S^2\le (2P + 3)\cdot S + 2P\ }$ . Therefore,

$\blacktriangleright\ \left\|\begin{array}{c} a^2\le a \\\ \ b^2\le b\end{array}\right\|$ $\implies$ $a^2 + b^2\le a + b$ $\implies$ $S^2 - S\le 2P$ , with equality iff $\{a,b\} = \{0,1\}$ .

$\blacktriangleright\ (1 - a)(1 - b)\ge 0$ $\implies$ $S\le P + 1$ $\implies$ $2S + 1\le 2P + 3$ , with equality iff $1\in\{a,b\}$ .

$\blacktriangleright\ 3S^2 = (2S + 1)\cdot S + (S^2 - S)\le (2P + 3)\cdot S + 2P$ $\implies$ $\boxed {\ 3S^2\le (2P + 3)\cdot S + 2P\ }$

with equality iff $\{a,b\} = \{0,1\}$ and $1\in \{a,b\}$ , i.e. $\left\|\begin{array}{c} a = 1 \\ \ b = 1\end{array}\right\|$ or $\left\|\begin{array}{c} a = 1 \\ \ b = 0\end{array}\right\|$ or $\left\|\begin{array}{c} a = 0 \\ \ b = 1\end{array}\right\|$ .

Lemma (Goldstone's inequality). Let $\triangle ABC$ with the circumcircle $\mathbb C(O,R)$ and the incircle $\mathbb C(I,r)\ .$ Prove that there is the inequality $\boxed {\ \frac 1{R^2}\le\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}\le\frac 1{4r^2}\ }$ (standard notations).

Proof.

$\bullet$ $\left\{\begin{array}{c} 4(s-a)(s-b)\le c^2\\\\ 4(s-a)(s-c)\le b^2\end{array}\right|\bigodot\implies$ $16(s-a)sr^2\le b^2c^2\iff$ $4(s-a)rabc\le b^2c^2R\iff$ $\frac {bc}{a}\ge \frac {4r(s-a)}{R}\implies$ $\sum \frac {bc}{a}\ge \frac {4rs}{R}\implies$ $\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}\ge\frac 1{R^2}\ (1)\ .$

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

In conclusion, from the relations $(1)$ and $(2)$ obtain that bilateral inequality $\boxed {\ \frac 1{R^2}\le\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}\le\frac 1{4r^2}\ }$ with the name Goldstone's inequality.

PP11 (George APOSTOLOPOULOS, Greece. Let $\triangle ABC$ with the circumcircle $\mathbb C(O,R)$ and the incircle $\mathbb C(I,r)\ .$ Prove that there is the inequality $\boxed {\ \frac 1{IA^2}+\frac 1{IB^2}+\frac 1{IC^2}\ge 3\cdot\left(\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}\right)\ }\ .$

Proof. We can prove easily or can use the well known identities $\boxed{\ \sum a(s-b)(s-c)=2sr(2R-r)\ }\ (1)$ and $\boxed{\ IA^2=\frac {bc(s-a)}s\ }\ (2)$ a.s.o. Therefore, $\frac 1{IA^2}\ \stackrel{(2)}{=}\ \frac s{bc(s-a)}=$ $\frac {a\cancel s}{4R\cancel sr(s-a)}=$

$\frac 1{4Rr}\cdot\frac a{s-a}$ $\implies$ $\frac 1{IA^2}=\frac 1{4Rr}\cdot\frac a{s-a}$ a.s.o. $\implies$ $\sum\frac 1{IA^2}=\frac 1{4Rr}\cdot\sum \frac a{s-a}=$ $\frac 1{4Rr}\cdot \frac {\sum a(s-b)(s-c)}{(s-a)(s-b)(s-c)}\ \stackrel{(1)}{=}\ \frac 1{4R\cancel r}\cdot \frac {2\cancel s\cancel r(2R-r)}{\cancel sr^2}=$ $\frac {2R-r}{2Rr^2}\implies$ $\boxed{\ \sum\frac 1{IA^2}=\frac {2R-r}{2Rr^2}\ }\ (3)\ .$

I"ll apply upper Goldstone's inequality $:\ \sum \frac 1{a^2}\le \frac 1{4r^2}\iff$ $3\cdot\sum\frac 1{a^2}\le\frac 3{4r^2}$ and $\frac 3{4r^2}\le \frac {2R-r}{2Rr^2}\ \stackrel{(3)}{=}\ \sum\frac 1{IA^2}$ because $\frac 3{4\cancel{r^2}}\le \frac {2R-r}{2R\cancel{r^2}}\iff$ $\frac 32\le \frac {2R-r}R\iff$ $2r\le R\ ,$ what is true.

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This post has been edited 201 times. Last edited by Virgil Nicula, Jan 30, 2018, 6:44 PM

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234. Some simple and nice problems from contests.

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

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232. Some geometrical inequalities (own).

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

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

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

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

173. An interesting trigonometric equations.

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

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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.

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

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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.

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

119. Inspired by a Miculita's problem.

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

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

111. Simple, but interesting ...

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101. O problema cu numere complexe.

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88. A nice geometric locus conditioned metrically.

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

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

81. Sawayama and Thebault’s theorem.

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

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60. Mediteranean M.C. 2010 Problem 3.

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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.

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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).

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21 bis. Some problems with complex numbers.

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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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