An All-Inclusive Inequality

Problem

An All-Inclusive Inequality

Proof 1

An All-Inclusive Inequality, Proof 1

$\displaystyle A'B=A'C; OA'\perp BC; O - \text{circumcenter} $

$\displaystyle AA'\leq A'O+OA $

$\displaystyle m_a\leq R\cos A+R=R\Bigr(2\cos^2 \frac{A}{2}-1+1\Bigr)=2R\cos^2 \frac{A}{2} $

$\displaystyle m_a\leq 2R\cos^2 \frac{A}{2}; m_b\leq 2R\cos^2 \frac{B}{2}; m_c\leq 2R\cos^2 \frac{C}{2} $

$\displaystyle m_am_bm_c\leq 8R^3\cos^2\frac{A}{2}\cos^2\frac{B}{2}\cos^2 \frac{C}{2} $

(By multiplying)

$\displaystyle m_am_bm_c\leq 8R^3\frac{s(s-a)}{bc}\cdot \frac{s(s-b)}{ac}\cdot \frac{s(s-c)}{ab} $

$\displaystyle m_am_bm_c\leq 8 \frac{R^3s^2S^2}{a^2b^2c^2}=\frac{8R^3s^2S^2}{16R^2S^2}= $

$\displaystyle =\frac{8Rs^2}{16}=\frac{Rs^2}{2}=\frac{s^2rR}{2r} $

(1)

$\displaystyle m_am_bm_c\leq \frac{s^2rR}{2r}.$


(2)

$\displaystyle r_ar_br_c=\frac{S^3}{(p-a)(p-b)(p-c)}=\frac{sS^3}{S^2}=rS^2$

By (1),(2):

$\displaystyle \frac{m_am_bm_c}{r_ar_br_c}\leq \frac{s^2rR}{2r}\cdot \frac{1}{rs^2}=\frac{R}{2r} $

(3)

$\displaystyle \frac{m_am_bm_c}{r_ar_br_c}\leq \frac{R}{2r}$

$\displaystyle \ell_a\ell_b\ell_c\leq s^2r \text{ (Carlitz's inequality American Mathematical Monthly - 1963)} $

$\displaystyle \frac{\ell_a\ell_b\ell_c}{h_ah_bh_c }\leq \frac{s^2r}{h_ah_bh_c}=\frac{s^2r}{\frac{2S}{a}\cdot \frac{2S}{b}\cdot \frac{2S}{c}}= $

$\displaystyle =\frac{s^2rabc}{8S^3}=\frac{s^2r\cdot 4RS}{8S^3}=\frac{RS\cdot s^2r}{2S^3}= $

$\displaystyle =\frac{R\cdot S^2s}{2S^3}=\frac{Rs}{2S}=\frac{Rs}{2rs}=\frac{R}{2r} $

(4)

$\displaystyle\frac{\ell_a\ell_b\ell_c}{h_ah_bh_c }\leq \frac{R}{2r}$

Adding (3),(4):

$\displaystyle \frac{m_am_bm_c}{r_ar_br_c}+\frac{\ell_a\ell_b\ell_c}{h_ah_bh_c}\leq \frac{R}{2r}+\frac{R}{2r}=\frac{R}{r} $

Proof 2

$\displaystyle \frac{\ell_a\ell_b\ell_c}{h_ah_bh_c}=\frac{\frac{2bc \cos \frac{A}{2}}{b+c}\cdot \frac{2ca \cos \frac{B}{2}}{c+a}\cdot \frac{2ab \cos \frac{C}{2}}{a+b}}{(\frac{bc}{2R})(\frac{ca}{2R})(\frac{ab}{2R})} $

$\displaystyle =\frac{64R^3\sqrt{\frac{s(s-a)}{bc}\cdot \frac{s(s-b)}{ca}\cdot \frac{s(s-c)}{ab}}}{(a+b)(b+c)(c+a)}=\frac{64R^3 s\cdot \frac{\Delta}{abc}}{(a+b)(b+c)(c+a)} $

$\displaystyle =\frac{64R^3s\frac{\Delta}{4R\Delta}}{(a+b)(b+c)(c+a)}=\frac{16R^2s}{(a+b)(b+c)(c+a)} $

$\displaystyle (a+b)(b+c)(c+a)=(2s-c)(2s-a)(2s-b)= $

$\displaystyle =8s^3-4s^2(2s)+2s\Bigr(\sum ab\Bigr)-abc= $

$\displaystyle =2s(s^2+4Rr+r^2)-4Rrs=2s(s^2+2Rr+r^2) $

$\displaystyle \frac{\ell_a\ell_b\ell_c}{h_ah_bh_c}=\frac{8R^2}{s^2+12Rr+r^2} $

$\displaystyle \text{Now, } r_ar_br_c=\frac{\Delta^2}{r}=\frac{r^2s^2}{r}=rs^2 $

Given inequality

$\displaystyle \Leftrightarrow \frac{m_am_bm_c}{rs^2}+\frac{8R^2}{s^2+2Rr+r^2}\leq \frac{R}{r} $

$\displaystyle \Leftrightarrow \frac{m_am_bm_c}{rs^2}\leq \frac{R}{r}-\frac{8R^2}{s^2+2Rr+r^2} $

$\displaystyle \Leftrightarrow m_am_bm_c\leq Rs^2-\frac{8R^2rs^2}{s^2+2Rr+r^2}= $

$\displaystyle =Rs^2\Bigr(1-\frac{8Rr}{s^2+2Rr+r^2}\Bigr)=\frac{Rs^2(s^2-6Rr+r^2)}{s^2+2Rr+r^2} $

(5)

$\displaystyle\Leftrightarrow m_a^2m_b^2m_c^2\leq \frac{R^2s^4(s^2-6Rr+r)^2}{(s^2+2Rr+r^2)^2}$

$\displaystyle m_a^2m_b^2m_c^2=\frac{1}{64}(2b^2+2c^2-a^2)(2c^2+2a^2-b^2)(2a^2+2b^2-c^2) $

$\displaystyle =\frac{1}{64}\Biggl\{-4\Bigr(\sum a^6\Bigr)+6\Bigr(\sum a^4b^2+\sum a^2b^4\Bigr)+3a^2b^2c^2\Bigr\} $

$\displaystyle \text{Now, } \sum a^6=\Bigr(\sum a^2\Bigr)^3-3(a^2+b^2)(b^2+c^2)(c^2+a^2) $

$\displaystyle (a^2+b^2)(b^2+c^2)(c^2+a^2)=\Bigr(\sum a^2-c^2\Bigr)\Bigr(\sum a^2-a^2\Bigr)\Bigr(\sum a^2-b^2\Bigr) $

$\displaystyle =\Bigr(a^2\Bigr)^3-\Bigr(\sum a^2\Bigr)^2\Bigr(\sum a^2\Bigr)+\Bigr(\sum a^2\Bigr)\Bigr(\sum a^2b^2\Bigr)-a^2b^2c^2 $

$\displaystyle =\Bigr(\sum a^2\Bigr)\Bigr(\sum a^2b^2\Bigr)-a^2b^2c^2 $

$\displaystyle \sum a^6=\Bigr(\sum a^2\Bigr)^3-3\Bigr(\sum a^2\Bigr)\Bigr(\sum a^2 b^2\Bigr)+3a^2b^2c^2 $

$\displaystyle \text{Also, } \sum a^4b^2+\sum a^2b^4= $

$\displaystyle =a^2b^2\Bigr(\sum a^2-c^2\Bigr)+b^2c^2\Bigr(\sum a^2-a^2\Bigr)+c^2a^2\Bigr(a^2-b^2\Bigr) $

$\displaystyle =\Bigr(\sum a^2\Bigr)\Bigr(\sum a^2b^2\Bigr)-3a^2b^2c^2 $

$\displaystyle\begin{align} m_a^2m_b^2m_c^2=\frac{1}{64}\Bigr\{&-4\Bigr(\sum a^2\Bigr)^3+12\Bigr(\sum a^2\Bigr)\Bigr(\sum a^2b^2\Bigr)-12a^2b^2c^2+6\Bigr(\sum a^2\Bigr)\Bigr(a^2b^2\Bigr)\\ &-18a^2b^2c^2+3a^2b^2c^2\Bigr\}\\ &=\frac{1}{32}\Biggl\{-2\Bigr(\sum a^2\Bigr)^3+9\Bigr(\sum a^2\Bigr)\Bigr(\sum a^2b^2\Bigr)-\frac{27}{2}a^2b^2c^2\Biggl\} \end{align}$

$\displaystyle\begin{align} \sum a^2b^2&=\Bigr(\sum ab\Bigr)^2-2abc(2s)\\ &=(s^2+4Rr+r^2)^2-16Rrs^2\\ &=s^4+16R^2r^2+r^4-8Rrs^2+8Rr^3+2s^2r^2 \end{align}$

$\begin{align}\displaystyle m_a^2m_b^2m_c^2=\frac{1}{32}\Bigr\{&-16(s^2-4Rr-r^2)^3+18(s^2-4Rr-r^2)\\&\cdot \Bigr(s^4+16R^2r^2+r^4-8Rrs^2+8Rr^3+2s^2r^2-\frac{27}{2}(4Rrs)^2\Bigr)\Bigr\} \end{align}$

(6)

$\begin{align}&=\frac{1}{16}(s^6-12s^4Rr+33s^4r^2-60s^2R^2r^2\\&-120s^2Rr^3-33s^2r^4-64R^3r^3-48R^2r^4-12Rr^5-r^6) \end{align}$

Given inequality:

$\displaystyle\begin{align}\Leftrightarrow&(s^6-12s^4Rr+33s^4r^2-60s^2R^2r^2\\ &-120s^2Rr^3-33s^2r^4-64R^3r^3-48R^2r^4-12Rr^5-r^6)\\ &\leq \frac{R^2s^4(s^2-6Rr+r^2)^2}{(s^2+2Rr+r^2)^2} \end{align}$

From (5),(6):

$\displaystyle \Leftrightarrow s^{10}+s^8(35r^2-8Rr-16R^2)+s^6(34r^4-136R^2r^2-8Rr^3+192R^3r)- $

$\displaystyle -s^4(160R^3r^3+576R^4r^2+580R^2r^4+264Rr^5+34r^6)- $

$\displaystyle -s^2(496R^4r^4+1040R^3r^5+816R^2r^6+280Rr^7+35r^8)-256R^5r^5- $

(A)

$-448R^4r^6-304R^3r^7-100R^2r^8-16Rr^9-r^{10}\leq 0$

$\displaystyle LHS \text{ of } (A) \leq (4R^2+4Rr+3r^2)^5+ $

$ +(4R^2+4Rr+3r^2)^4(35r^2-8Rr-16R^2)+ $

$ +(4R^2+4Rr+3r^2)^3(34r^4-136R^2r^2-8Rr^3+192R^3r)- $

$\displaystyle -(16Rr-5r^2)^2(160R^3r^3+576R^4r^2+580R^2r^4+264Rr^5+34r^6)- $

$\displaystyle -(16Rr-5r^2)(496R^4r^4+1040R^3r^5+816R^2r^6+280Rr^7+35r^8)- $

$\displaystyle -256R^5r^5-448R^4r^6-304R^3r^7-100R^2r^8-16Rr^9-r^{10}= $

$\displaystyle $

$\displaystyle =-3072R^{10}-1024R^9r+6144R^8r^2+27648R^7r^3-91776R^6r^4+ $

$\displaystyle +130176R^5r^5-52240R^4r^6+97984R^3r^7+69116R^2r^8+19212Rr^9+3320r^{10} $

It suffices to prove:

$\displaystyle -3072R^{10}-1024R^9r+6144R^8r^2+27648R^7r^3-91776R^6r^4+130176R^5r^5- $

$\displaystyle -52240R^4r^6+97984R^3r^7+69116R^2r^8+19212Rr^9+3320r^{10}\leq 0 $

$\displaystyle \Leftrightarrow 3072t^{10}+1024t^9-6144t^8-27648t^7+91776t^6-130176t^5+ $

$\displaystyle +52240t^4-97984t^3-69116t^2-19212t-3320\geq 0 \Bigr(t=\frac{R}{r}\Bigr) $

$\displaystyle \Leftrightarrow 768t^{10}+256t^9-153t^8-6912t^7+22944t^6-32544t^5+13060t^4- $

$\displaystyle -24496t^3-17279t^2-4803t-830\geq 0 $

(B)

$\begin{align}\Leftrightarrow (t-2)(768t^9&+1792t^8+2048t^7-2816t^6+17312t^5\\ &+2080t^4+17220t^3+9944t^2+2609t+415)\geq 0 \end{align}$

$\displaystyle \text{Now, } 768t^9-2816t^6+17220t^3=t^3(768t^6-2816t^3+17220) $

$\displaystyle =t^3\underbrace{\begin{matrix} (768z^2-2816z+17220)\\ e \end{matrix}}_{\Delta =-44969984<0 \quad e>0}(z=t^3) $

$\displaystyle 768t^9-2816t^6+17220t^3>0 $

$\displaystyle \Rightarrow (B) \text{ is true } (t=\frac{R}{r}\geq 2\to \text{Euler}) $

$\displaystyle \Rightarrow (A) \text{ is true (Hence proved)} $

Proof 3

$\displaystyle\begin{align} \prod_{cycl}\ell_a &= \prod_{cycl}\frac{2\sqrt{bc(p(p-a)}}{b+c}\\ &\le\prod_{cycl}\sqrt{p(p-a)}\\ &=\prod_{cycl}\sqrt{r_br_c}\\ &=r_ar_br_c=\frac{S^2}{r}=p^2r. \end{align}$

$\displaystyle \prod_{cycl}h_a = \prod_{cycl}\frac{2S}{a}=\frac{8S^3}{4SR}=\frac{2p^2r^2}{R}.$

It follows that

$\displaystyle\frac{\ell_a\ell_b\ell_c}{h_ah_bh_c}\le\frac{p^2r}{2p^2r^2/R}=\frac{R}{2r}.$

On the other hand,

$m_a\le R(1+\cos A),\;m_b\le R(1+\cos B),\;m_c\le R(1+\cos C)$

so that

$\displaystyle m_am_bm_c \le 8R^3\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}=\frac{Rp^2}{2},$

since

$\displaystyle\begin{align} p&=\frac{a+b+c}{2}=R(\sin A+\sin B+\sin C)\\ &=4R\cos\frac{A}{2}\cos\frac{A}{2}\cos\frac{A}{2}. \end{align}$

Also, $r_ar_br_c=rp^2,\;$ such that $\displaystyle\frac{m_am_bm_c}{r_ar_br_c}\le\frac{R}{2r}.\;$ And, finally,

$\displaystyle\frac{m_am_bm_c}{r_ar_br_c}+\frac{\ell_a\ell_b\ell_c}{h_ah_bh_c}\le\frac{R}{2r} +\frac{R}{2r}\le\frac{R}{r}.$

Acknowledgment

The problem has been kindly posted by Dan Sitaru, along with a solution (Proof 1), at the CutTheKnotMath facebook page; Proof 2 is by Soumava Chakraborty (India); Proof 3 is by Myagmarsuren Yadamsuren. I am greatly indebted to Dan Sitaru for supplying the LaTeX files.

 

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