Bugi wrote: For an acute triangle $ ABC$ prove the inequality: $ \sum_{cyclic} \frac {m_a^2}{ - a^2 + b^2 + c^2}\ge \frac {9}{4}$ where $ m_a,m_b,m_c$ are lengths of corresponding medians. \[ \Leftrightarrow \sum {\frac{{2b^2 + 2c^2 - a^2 }}{{b^2 + c^2 - a^2 }}} \ge 9 \] \[ \Leftrightarrow \sum {\frac{{b^2 + c^2 }}{{b^2 + c^2 - a^2 }}} \ge 6 \] \[ \Leftrightarrow \sum {\frac{{a^2 }}{{b^2 + c^2 - a^2 }}} \ge 3 \] which is always true since \[ \sum {\frac{{a^2 }}{{b^2 + c^2 - a^2 }}} \ge 3\sqrt[3]{{\frac{{a^2 b^2 c^2 }}{{\prod {\left( {b^2 + c^2 - a^2 } \right)} }}}} \ge 3 \]

How did you get $ 2b^2+2c^2-a^2$ in the numerator?

It's a well known fact: $ m_a^2 = \frac {2b^2 + 2c^2 - a^2}{4}$ and without it, most of the inequalities in which medians figure aren't solvable. See here, for instance (equation(1)): http://mathworld.wolfram.com/TriangleMedian.html

Bugi wrote: For an acute triangle $ ABC$ prove the inequality: $ \sum_{cyclic} \frac {m_a^2}{ - a^2 + b^2 + c^2}\ge \frac {9}{4}$ where $ m_a,m_b,m_c$ are lengths of corresponding medians. $ (m_a)^2 = (\frac {a}{2})^2 + c^2 - ac*cos(B)$ By the cosine rule. Also by the cosine rule: $ cos(B) = \frac {a^2 + c^2 - b^2}{2ac}$ so we get $ (m_a)^2 = \frac {2b^2 + 2c^2 - a^2}{4}$ Thus we get: $ \sum_{cyc} \frac {2b^2 + 2c^2 - a^2}{b^2 + c^2 - a^2} \ge 9$. Since $ \triangle ABC$ is acute we know that $ cos(C) > 0$ And thus using $ a^2 + b^2 - 2ab.cos(C) = c^2$ But $ a^2 + b^2 - 2ab.cos(C) > a^2 + b^2$ and thus $ c^2 > a^2 + b^2$. So $ a^2,b^2,c^2$ are sides in a triangle. Thus we set $ a^2 = x + y$, $ b^2 = y + z$, $ c^2 = z + x$ with $ x,y,z > 0$. So we just have: $ \sum_{cyc} \frac {4z + x + y}{2y} \ge 9$, which follows directly by $ \sum_{cyc} \frac {4z}{2y} \ge 6$, $ \sum_{cyc} \frac {x}{2y} \ge \frac {3}{2}$, $ \sum_{cyc} \frac {y}{2y} = \frac {3}{2}$. (By AM-GM)

hsiljak wrote: It's a well known fact: $ m_a^2 = \frac {2b^2 + 2c^2 - a^2}{4}$ and without it, most of the inequalities in which medians figure aren't solvable. See here, for instance (equation(1)): http://mathworld.wolfram.com/TriangleMedian.html The equality is true by Stewart therorem

Honey_S wrote: $ \Leftrightarrow \sum {\frac {{b^2 + c^2 }}{{b^2 + c^2 - a^2 }}} \ge 6$ Here is how I proved this. Let $ x = a^2$, $ y = b^2$, $ z = c^2$ So we have to show $ \sum \frac {x + y}{x + y - z} \ge 6$. This fraction is $ > 0$, since $ x + y > z$. This is equivalent to $ \sum \frac {1}{1 - \frac {z}{x + y}} \ge 6$. By Cauchy, $ \sum \frac {1}{1 - \frac {z}{x + y}} \ge \frac {9}{3 - \sum \frac {z}{x + y}}$. But by Nesbitt's inequality, we know $ \sum \frac {z}{x + y} \ge \frac {3}{2}$, which completes the proof. By the way, how do you prove Honey_S's last inequality, that $ xyz \ge (x + y - z)(x - y + z)( - x + y + z)$? EDIT: Now I remember. This is either Schur, or you can use that $ (x+(y-z))(x-(y-z))=x^2-(y-z)^2 \le x^2$, and multiply it cyclically, which proves a case of Schur.

...or you can substitute $ x+y-z=a;y+z-x=b;z+x-y=c $ with $ x+y+z=a+b+c $ and your last inequality (dgreenb801) is now equivalent with $ (a+b)(b+c)(c+a) \ge 8abc $, which is true by $AM-GM$. Note that $ a,b,c $ are positive since $ x+y>z, y+z>x, z+x>y $ by triangle inequality, so you are allowed to apply the $AM-GM$ inequality.

For an acute triangle $ ABC$ prove the inequality: $ \sum_{cyclic} \frac{h_a^2}{-a^2+b^2+c^2}\ge \frac{9}{4}$ where $h_a=\frac{2S}{a}$. Because; $Q={\frac {6}{11}}\, \left( -{s}^{2}+ \left( -4\,R+8\,r \right) s+12\,{ R}^{2}-5\,{r}^{2}-8\,Rr \right) {R}^{2}+{\frac {92}{33}}\, \left( -{s} ^{2}+ \left( -4\,r+2\,R \right) s+r \left( 7\,r+10\,R \right) \right) {r}^{2}+ \left( \left( 4\,{R}^{2}+4\,Rr+3\,{r}^{2}-{s}^{2} \right) \left( R-r \right) -{r}^{2} \left( R-2\,r \right) \right) \left( {\frac {130}{11}}\,R+{\frac {196}{11}}\,r \right) +{\frac {52} {33}}\, \left( -{s}^{2}+ \left( -4\,r+2\,R \right) s+r \left( -5\,r+16 \,R \right) \right) {r}^{2}+2\, \left( R-2\,r \right) ^{2}{r}^{2}+{ \frac {41}{99}}\, \left( -{s}^{2}+ \left( -4\,r+2\,R \right) s+r \left( 7\,r+10\,R \right) \right) ^{2}+{\frac {59}{99}}\, \left( -{s }^{2}+ \left( -4\,r+2\,R \right) s+r \left( 7\,r+10\,R \right) \right) \left( -{s}^{2}+ \left( -4\,R+8\,r \right) s+12\,{R}^{2}-5\, {r}^{2}-8\,Rr \right) +{\frac {3}{22}}\, \left( {s}^{2}-16\,Rr+5\,{r}^ {2} \right) \left( -{s}^{2}+ \left( -4\,R+8\,r \right) s+12\,{R}^{2}- 5\,{r}^{2}-8\,Rr \right) +{\frac {25}{198}}\, \left( -{s}^{2}+ \left( -4\,R+8\,r \right) s+12\,{R}^{2}-5\,{r}^{2}-8\,Rr \right) ^{2}\geq 0. $ ${\frac {{{\it ha}}^{2}}{-{a}^{2}+{b}^{2}+{c}^{2}}}+{\frac {{{\it hb}}^ {2}}{-{b}^{2}+{c}^{2}+{a}^{2}}}+{\frac {{{\it hc}}^{2}}{-{c}^{2}+{a}^{ 2}+{b}^{2}}}-9/4=4\,{\frac {{s}^{2}{r}^{2}Q}{{R}^{2} \left( -{a}^{2}+{ b}^{2}+{c}^{2} \right) \left( -{b}^{2}+{c}^{2}+{a}^{2} \right) \left( -{c}^{2}+{a}^{2}+{b}^{2} \right) }} $ BQ

Or this: In any triangle,prove that; ${{\it ha}}^{2} \left( -{b}^{2}+{a}^{2}+{c}^{2} \right) \left( {b}^{2} +{a}^{2}-{c}^{2} \right) +{{\it hb}}^{2} \left( {b}^{2}+{a}^{2}-{c}^{2 } \right) \left( {b}^{2}+{c}^{2}-{a}^{2} \right) +{{\it hc}}^{2} \left( {b}^{2}+{c}^{2}-{a}^{2} \right) \left( -{b}^{2}+{a}^{2}+{c}^{ 2} \right) \geq 9/4\, \left( {b}^{2}+{a}^{2}-{c}^{2} \right) \left( {b}^{ 2}+{c}^{2}-{a}^{2} \right) \left( -{b}^{2}+{a}^{2}+{c}^{2} \right) $ ha=2S/a. BQ

xzlbq wrote:

Or this: In any triangle,prove that; ${{\it ha}}^{2} \left( -{b}^{2}+{a}^{2}+{c}^{2} \right) \left( {b}^{2} +{a}^{2}-{c}^{2} \right) +{{\it hb}}^{2} \left( {b}^{2}+{a}^{2}-{c}^{2 } \right) \left( {b}^{2}+{c}^{2}-{a}^{2} \right) +{{\it hc}}^{2} \left( {b}^{2}+{c}^{2}-{a}^{2} \right) \left( -{b}^{2}+{a}^{2}+{c}^{ 2} \right) =9/4\, \left( {b}^{2}+{a}^{2}-{c}^{2} \right) \left( {b}^{ 2}+{c}^{2}-{a}^{2} \right) \left( -{b}^{2}+{a}^{2}+{c}^{2} \right) $ ha=2S/a. BQ xzlbq wrote:

Or this: In any triangle,prove that; ${{\it ha}}^{2} \left( -{b}^{2}+{a}^{2}+{c}^{2} \right) \left( {b}^{2} +{a}^{2}-{c}^{2} \right) +{{\it hb}}^{2} \left( {b}^{2}+{a}^{2}-{c}^{2 } \right) \left( {b}^{2}+{c}^{2}-{a}^{2} \right) +{{\it hc}}^{2} \left( {b}^{2}+{c}^{2}-{a}^{2} \right) \left( -{b}^{2}+{a}^{2}+{c}^{ 2} \right) \geq 9/4\, \left( {b}^{2}+{a}^{2}-{c}^{2} \right) \left( {b}^{ 2}+{c}^{2}-{a}^{2} \right) \left( -{b}^{2}+{a}^{2}+{c}^{2} \right) $ ha=2S/a. BQ Who can do The ineq ,in others? It is stronger and nice ineqtility. BQ

But nice and hard is this: In any triangle,prove that: ${\it mb}\,{\it mc}\, \left( -{b}^{2}+{a}^{2}+{c}^{2} \right) \left( { b}^{2}+{a}^{2}-{c}^{2} \right) +{\it mc}\,{\it ma}\, \left( {b}^{2}+{a }^{2}-{c}^{2} \right) \left( {b}^{2}+{c}^{2}-{a}^{2} \right) +{\it ma }\,{\it mb}\, \left( {b}^{2}+{c}^{2}-{a}^{2} \right) \left( -{b}^{2}+ {a}^{2}+{c}^{2} \right) \geq 9/4\, \left( {b}^{2}+{a}^{2}-{c}^{2} \right) \left( {b}^{2}+{c}^{2}-{a}^{2} \right) \left( -{b}^{2}+{a}^{2}+{c}^{ 2} \right) $ if get $h_a,h_b,h_c$,easy BQ

In any triangle,prove that: ${\frac {{\it ha}}{s-a}}+{\frac {{\it hb}}{s-b}}+{\frac {{\it hc}}{s-c} }>5. $

xzlbq wrote: In any triangle,prove that: ${\frac {{\it ha}}{s-a}}+{\frac {{\it hb}}{s-b}}+{\frac {{\it hc}}{s-c} }>5. $ s means a+b+c/2 , right?

mathmatecS wrote: xzlbq wrote: In any triangle,prove that: ${\frac {{\it ha}}{s-a}}+{\frac {{\it hb}}{s-b}}+{\frac {{\it hc}}{s-c} }>5. $ s means a+b+c/2 , right? Yes. BQ

${\frac {{{\it ha}}^{2}}{ \left( s-a \right) ^{2}}}+{\frac {{{\it hb}}^ {2}}{ \left( s-b \right) ^{2}}}+{\frac {{{\it hc}}^{2}}{ \left( s-c \right) ^{2}}}-9=1/4\,{\frac {Q}{{s}^{2}{R}^{2}}}$ $Q= \left( R-2\,r \right) ^{2} \left( 4\,{r}^{2}+16\,Rr \right) + \left( 4\,{R}^{2}+4\,Rr+3\,{r}^{2}-{s}^{2} \right) ^{2}+ \left( R-2\, r \right) \left( 4\,{R}^{2}+4\,Rr+3\,{r}^{2}-{s}^{2} \right) \left( 6\,R+4\,r \right) +54\, \left( 4\,{R}^{2}+4\,Rr+3\,{r}^{2}-{s}^{2} \right) {R}^{2}\geq 0. $ So ${\frac {{{\it ha}}^{2}}{ \left( s-a \right) ^{2}}}+{\frac {{{\it hb}}^ {2}}{ \left( s-b \right) ^{2}}}+{\frac {{{\it hc}}^{2}}{ \left( s-c \right) ^{2}}}\geq 9$

xzlbq wrote: ${\frac {{{\it ha}}^{2}}{ \left( s-a \right) ^{2}}}+{\frac {{{\it hb}}^ {2}}{ \left( s-b \right) ^{2}}}+{\frac {{{\it hc}}^{2}}{ \left( s-c \right) ^{2}}}-9=1/4\,{\frac {Q}{{s}^{2}{R}^{2}}}$ $Q= \left( R-2\,r \right) ^{2} \left( 4\,{r}^{2}+16\,Rr \right) + \left( 4\,{R}^{2}+4\,Rr+3\,{r}^{2}-{s}^{2} \right) ^{2}+ \left( R-2\, r \right) \left( 4\,{R}^{2}+4\,Rr+3\,{r}^{2}-{s}^{2} \right) \left( 6\,R+4\,r \right) +54\, \left( 4\,{R}^{2}+4\,Rr+3\,{r}^{2}-{s}^{2} \right) {R}^{2}\geq 0. $ So ${\frac {{{\it ha}}^{2}}{ \left( s-a \right) ^{2}}}+{\frac {{{\it hb}}^ {2}}{ \left( s-b \right) ^{2}}}+{\frac {{{\it hc}}^{2}}{ \left( s-c \right) ^{2}}}\geq 9$ Oh I didin't know what h means so...

$h_a=\frac{2S}{a},$ $S=\sqrt{s(s-a)(s-b)(s-c)}$ $s=1/2(a+b+c)$ BQ

Honey_S wrote: which is always true since \[ \sum {\frac{{a^2 }}{{b^2 + c^2 - a^2 }}} \ge 3\sqrt[3]{{\frac{{a^2 b^2 c^2 }}{{\prod {\left( {b^2 + c^2 - a^2 } \right)} }}}} \ge 3 \] why it's true?

xzlbq wrote: But nice and hard is this: In any triangle,prove that: ${\it mb}\,{\it mc}\, \left( -{b}^{2}+{a}^{2}+{c}^{2} \right) \left( { b}^{2}+{a}^{2}-{c}^{2} \right) +{\it mc}\,{\it ma}\, \left( {b}^{2}+{a }^{2}-{c}^{2} \right) \left( {b}^{2}+{c}^{2}-{a}^{2} \right) +{\it ma }\,{\it mb}\, \left( {b}^{2}+{c}^{2}-{a}^{2} \right) \left( -{b}^{2}+ {a}^{2}+{c}^{2} \right) \geq 9/4\, \left( {b}^{2}+{a}^{2}-{c}^{2} \right) \left( {b}^{2}+{c}^{2}-{a}^{2} \right) \left( -{b}^{2}+{a}^{2}+{c}^{ 2} \right) $ if get $h_a,h_b,h_c$,easy BQ with Klmkin to it,get this: ${\frac {9}{16}}\,bc \left( -{{\it mb}}^{2}+{{\it ma}}^{2}+{{\it mc}}^{ 2} \right) \left( {{\it mb}}^{2}+{{\it ma}}^{2}-{{\it mc}}^{2} \right) +{\frac {9}{16}}\,ca \left( {{\it mb}}^{2}+{{\it ma}}^{2}-{{ \it mc}}^{2} \right) \left( {{\it mb}}^{2}+{{\it mc}}^{2}-{{\it ma}}^ {2} \right) +{\frac {9}{16}}\,ab \left( {{\it mb}}^{2}+{{\it mc}}^{2}- {{\it ma}}^{2} \right) \left( -{{\it mb}}^{2}+{{\it ma}}^{2}+{{\it mc }}^{2} \right) -9/4\, \left( {{\it mb}}^{2}+{{\it ma}}^{2}-{{\it mc}}^ {2} \right) \left( {{\it mb}}^{2}+{{\it mc}}^{2}-{{\it ma}}^{2} \right) \left( -{{\it mb}}^{2}+{{\it ma}}^{2}+{{\it mc}}^{2} \right) \geq 0$ <=> $P={\frac {27}{16}}\,{s}^{6}-{\frac {45}{4}}\,r \left( 2\,R-r \right) { s}^{4}-9/4\,{r}^{2} \left( 31\,{R}^{2}+5\,Rr+{r}^{2} \right) {s}^{2}-{ \frac {9}{16}}\,{r}^{3} \left( 4\,R+r \right) ^{3}\geq$ $P={\frac {45}{16}}\, \left( R-2\,r \right) \left( {s}^{2}-16\,Rr+5\,{ r}^{2} \right) r{s}^{2}+{\frac {18909}{16}}\, \left( R-2\,r \right) {r }^{5}+603\, \left( R-2\,r \right) ^{2}{r}^{4}+{\frac {15021}{16}}\, \left( \left( {s}^{2}-16\,Rr+5\,{r}^{2} \right) \left( R-r \right) -{r}^{2} \left( R-2\,r \right) \right) {r}^{3}+ \left( {s}^{2}-16\,Rr +5\,{r}^{2} \right) ^{2} \left( {\frac {27}{16}}\,{s}^{2}+{\frac {459} {16}}\,Rr \right) +{\frac {1845}{4}}\, \left( R-2\,r \right) \left( \left( {s}^{2}-16\,Rr+5\,{r}^{2} \right) \left( R-r \right) -{r}^{2} \left( R-2\,r \right) \right) {r}^{2}+{\frac {1251}{4}}\, \left( {s} ^{2}-16\,Rr+5\,{r}^{2} \right) R{r}^{3}$

By Stewart's Theorem, we have $m_a=\frac{b^2+c^2-\frac{1}{2}a^2}{2}$, so we have \begin{align*} \sum_{cyc}\frac{m_a^2}{-a^2+b^2+c^2}\geq \frac{9}{4} &\iff \sum_{cyc} \frac{b^2+c^2}{-a^2+b^2+c^2} \geq 6\\ &\iff \sum_{cyc} \frac{a^2}{b^2+c^2-a^2}\geq 3 \\ &\iff \sum_{cyc} \frac{b^2+c^2-2bc\cos A}{2bc \cos A}\geq 3 \\ &\iff \sum_{cyc} \frac{b^2+c^2}{2bc \cos A}\geq 6 \\ &\iff \sum_{cyc} \sec A \geq 6 \end{align*}where we used AM-GM in the second last step to show that $b^2+c^2\geq 2bc$. Now, notice that since triangle $ABC$ is acute, we have $f(x)=\sec x$ is convex on the interval $(0,\frac{\pi}{2})$ as $f''(x)=\sec x(\sec^2 x + \tan^2 x) > 0$. So, by Jensen's Inequality, we have $\sum_{cyc} \sec A\geq 3\cdot \sec {\frac{\pi}{3}}= \boxed{6}$ .

MATH1945 wrote: Honey_S wrote: which is always true since \[ \sum {\frac{{a^2 }}{{b^2 + c^2 - a^2 }}} \ge 3\sqrt[3]{{\frac{{a^2 b^2 c^2 }}{{\prod {\left( {b^2 + c^2 - a^2 } \right)} }}}} \ge 3 \] why it's true? It's AM-GM and Schur

xzlbq wrote: For an acute triangle $ ABC$ prove the inequality: $ \sum_{cyclic} \frac{h_a^2}{-a^2+b^2+c^2}\ge \frac{9}{4}$ where $h_a=\frac{2S}{a}$. Because; $Q={\frac {6}{11}}\, \left( -{s}^{2}+ \left( -4\,R+8\,r \right) s+12\,{ R}^{2}-5\,{r}^{2}-8\,Rr \right) {R}^{2}+{\frac {92}{33}}\, \left( -{s} ^{2}+ \left( -4\,r+2\,R \right) s+r \left( 7\,r+10\,R \right) \right) {r}^{2}+ \left( \left( 4\,{R}^{2}+4\,Rr+3\,{r}^{2}-{s}^{2} \right) \left( R-r \right) -{r}^{2} \left( R-2\,r \right) \right) \left( {\frac {130}{11}}\,R+{\frac {196}{11}}\,r \right) +{\frac {52} {33}}\, \left( -{s}^{2}+ \left( -4\,r+2\,R \right) s+r \left( -5\,r+16 \,R \right) \right) {r}^{2}+2\, \left( R-2\,r \right) ^{2}{r}^{2}+{ \frac {41}{99}}\, \left( -{s}^{2}+ \left( -4\,r+2\,R \right) s+r \left( 7\,r+10\,R \right) \right) ^{2}+{\frac {59}{99}}\, \left( -{s }^{2}+ \left( -4\,r+2\,R \right) s+r \left( 7\,r+10\,R \right) \right) \left( -{s}^{2}+ \left( -4\,R+8\,r \right) s+12\,{R}^{2}-5\, {r}^{2}-8\,Rr \right) +{\frac {3}{22}}\, \left( {s}^{2}-16\,Rr+5\,{r}^ {2} \right) \left( -{s}^{2}+ \left( -4\,R+8\,r \right) s+12\,{R}^{2}- 5\,{r}^{2}-8\,Rr \right) +{\frac {25}{198}}\, \left( -{s}^{2}+ \left( -4\,R+8\,r \right) s+12\,{R}^{2}-5\,{r}^{2}-8\,Rr \right) ^{2}\geq 0. $ ${\frac {{{\it ha}}^{2}}{-{a}^{2}+{b}^{2}+{c}^{2}}}+{\frac {{{\it hb}}^ {2}}{-{b}^{2}+{c}^{2}+{a}^{2}}}+{\frac {{{\it hc}}^{2}}{-{c}^{2}+{a}^{ 2}+{b}^{2}}}-9/4=4\,{\frac {{s}^{2}{r}^{2}Q}{{R}^{2} \left( -{a}^{2}+{ b}^{2}+{c}^{2} \right) \left( -{b}^{2}+{c}^{2}+{a}^{2} \right) \left( -{c}^{2}+{a}^{2}+{b}^{2} \right) }} $ BQ $ \sum_{cyclic} \frac{h_a^2}{-a^2+b^2+c^2}\ge \frac{9}{4}$ where $h_a=\frac{2S}{a}$. <=> \[ \left( -{c}^{4}-{a}^{4}-{b}^{4}+2\,{a}^{2}{b}^{2}+2\,{a}^{2}{c}^{2}+2 \,{b}^{2}{c}^{2} \right) {\it \sum} \left( {\frac {1}{{a}^{2} \left( { b}^{2}+{c}^{2}-{a}^{2} \right) }} \right) \geq 9\]Let $[a=1/2\,\sqrt {2\,y+2\,z},b=1/2\,\sqrt {2\,z+2\,x},c=1/2\,\sqrt {2\,x+ 2\,y}] $ <=> \[ \left( xy+xz+yz \right) {\it \sum} \left( {\frac {1}{x \left( y+z \right) }} \right) \geq \frac{9}{2}\]Done!

But stronger is: In acute triangle,prove that \[{\frac {{\it h_a}}{\sqrt {{b}^{2}+{c}^{2}-{a}^{2}}}}+{\frac {{\it h_b}}{ \sqrt {{c}^{2}+{a}^{2}-{b}^{2}}}}+{\frac {{\it h_c}}{\sqrt {{b}^{2}+{a} ^{2}-{c}^{2}}}}\geq \frac{3}{2}\,\sqrt {3}\]

xzlbq wrote: But stronger is: In acute triangle,prove that \[{\frac {{\it h_a}}{\sqrt {{b}^{2}+{c}^{2}-{a}^{2}}}}+{\frac {{\it h_b}}{ \sqrt {{c}^{2}+{a}^{2}-{b}^{2}}}}+{\frac {{\it h_c}}{\sqrt {{b}^{2}+{a} ^{2}-{c}^{2}}}}\geq \frac{3}{2}\,\sqrt {3}\] <=> $x,y,z>0,$ \[\sqrt {xy+xz+yz} \left( {\frac {1}{\sqrt {x \left( y+z \right) }}}+{ \frac {1}{\sqrt {y \left( z+x \right) }}}+{\frac {1}{\sqrt {z \left( x +y \right) }}} \right) \geq \frac{3}{2}\,\sqrt {6}\]

In any triangle,,prove \[{{\it h_a}}^{2}{{\it h_b}}^{2}{{\it h_c}}^{2}\geq {\frac {27}{64}}\, \left( { b}^{2}+{c}^{2}-{a}^{2} \right) \left( {c}^{2}+{a}^{2}-{b}^{2} \right) \left( {b}^{2}+{a}^{2}-{c}^{2} \right) \]