has no solutions in integers 
,
,
.
be a 
board. Some of its cells are colored black in such a way that every 
board of 
black cells. Find the maximum amount of black cells that the board may have.
intersect at 
,
tangent to the extensions of 
at 
respectively. Circle 
passes through points 
,
.
.
let 
be the foot of the altitude from 
to the side 
and 
,
,
be the incenter, 
-
and 
the other intersection points of the circle 
with the lines 
and 
,
.
Y by yumeidesu, dm_edogawasonichi, arqady, younesmath2012maroc, Kunihiko_Chikaya, buratinogigle, livetolove212, linqaszayi, Skravin, Mateescu Constantin, ineX, trumpeter, baopbc, bgn, fractals, Hermitianism, anhtaitran, only-math, superpi, yyyshuai, Olympiad5642, bel.jad5, BG71, Nguyenhuyen_AG, kiyoras_2001, math_perfectionist, nguyenducmanh2705, Adventure10, Mango247
About four years ago, I began to research the geometric inequalities and recognize it as interesting as the algebraic inequalities. Here are some nice inequalities I think of, although they look nice and simple but I think some inequalities are very hard. Now, if you like geometric inequalities, you should try to prove them. Have fun! 
Given a triangle
with side lengths 
area 
circumradius 
inradius 
and any point 
. Let altitude lengths 
median lengths 
internal angle bisector lengths 
exradius lengths 
drawn from vertex 
and 
respectively. Prove that:
![$(2)\ \ \ \dfrac{1}{l_a}+\dfrac{1}{l_b}+\dfrac{1}{l_c} \ge \dfrac{3}{\sqrt[4]{3S^2}} .$](//latex.artofproblemsolving.com/0/7/3/07375a96903063ea354e65e355df9f8c3e1cd5db.png)
![$(11)\ \ \ \dfrac{9R}{2} \ge \dfrac{m_am_b}{m_c}+\dfrac{m_cm_a}{m_b}+\dfrac{m_bm_c}{m_a} \geq \sqrt[3]{9\big( m_a^3+m_b^3+m_c^3\big)} $](//latex.artofproblemsolving.com/e/9/1/e91d4bbce7060df3b31bcffd9b86f80f51c4ddeb.png)
.
For acute triangle
prove that
Given a triangle
with side lengths 
area 
circumradius 
inradius 
and any point 
. Let altitude lengths 
median lengths 
internal angle bisector lengths 
exradius lengths 
drawn from vertex 
and 
respectively. Prove that:
![$(2)\ \ \ \dfrac{1}{l_a}+\dfrac{1}{l_b}+\dfrac{1}{l_c} \ge \dfrac{3}{\sqrt[4]{3S^2}} .$](http://latex.artofproblemsolving.com/0/7/3/07375a96903063ea354e65e355df9f8c3e1cd5db.png)
![$(11)\ \ \ \dfrac{9R}{2} \ge \dfrac{m_am_b}{m_c}+\dfrac{m_cm_a}{m_b}+\dfrac{m_bm_c}{m_a} \geq \sqrt[3]{9\big( m_a^3+m_b^3+m_c^3\big)} $](http://latex.artofproblemsolving.com/e/9/1/e91d4bbce7060df3b31bcffd9b86f80f51c4ddeb.png)
.
For acute triangle
prove that 
This post has been edited 1 time. Last edited by quykhtn-qa1, Aug 7, 2016, 12:39 AM
Reason: edit latex
Reason: edit latex
are positive real numbers such that 
then![\[ \sqrt{\dfrac{x+1}{x+y}}+\sqrt{\dfrac{y+1}{y+z}}+\sqrt{\dfrac{z+1}{z+x}} \geq 3.\]](http://latex.artofproblemsolving.com/d/0/2/d0272e05378c8d3d7680cedfeaab0fe30087cddc.png)
,
and 
.
.
.
,
![\[(m_a+m_b+m_c)(\frac{1}{l_a}+\frac{1}{l_b}+\frac{1}{l_c})\ge (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\]](http://latex.artofproblemsolving.com/a/c/1/ac15f7de5c6c9ea81106f4c527c3f892a302b6ed.png)
![\[4m_al_a-(a+b+c)(-a+b+c)=\frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)(b-c)^2}{(b+c)^2(4m_al_a+(a+b+c)(-a+b+c))}\ge0\]](http://latex.artofproblemsolving.com/d/c/7/dc73c2b59ae511941b673b61905f88098afdd906.png)
![\[\frac{m_a}{l_a}\ge \frac{(a+b+c)(-a+b+c)}{4{l_a}^2}=\frac{(b+c)^2}{4bc}\]](http://latex.artofproblemsolving.com/5/1/6/516d564b459dc11f69ea328291444f1cde790dc5.png)
![\[(m_a+m_b+m_c)(\frac{1}{l_a}+\frac{1}{l_b}+\frac{1}{l_c})\ge \frac{\sum_{sym}a^2b+6abc}{4abc}+\sum_{cyc}\frac{m_bm_c}{l_a}=\]](http://latex.artofproblemsolving.com/9/7/8/978ed860d16b63b5f5605e0c3ff26773ee653aba.png)
![\[=\frac{\sum_{sym}a^2b+6abc}{4abc}+\frac{1}{l_al_bl_c}\sum_{cyc}l_a(l_bm_b+l_cmc)\ge \]](http://latex.artofproblemsolving.com/b/1/1/b1165c7de098aeea618ec2d154780e28f752afc9.png)
![\[\ge\frac{\sum_{sym}a^2b+6abc}{4abc}+\frac{1}{l_al_bl_c}\frac{1}{4}(a+b+c)\sum_{cyc}l_a((a-b+c)+(a+b-c))=\]](http://latex.artofproblemsolving.com/9/8/8/9883b0d0b44a77a9f80c240fd1f16b57bff8a34e.png)
![\[=\frac{\sum_{sym}a^2b+6abc}{4abc}+ \frac{1}{l_al_bl_c}\frac{1}{2}(a+b+c)\sum_{cyc}l_aa\]](http://latex.artofproblemsolving.com/a/9/e/a9e9f1149f62336f1ac4cef0744e58189a0bc152.png)
![\[2(a+b+c)\sum_{cyc}al_a\ge \frac{3(a+b)(b+c)(c+a)}{abc}l_al_bl_c\]](http://latex.artofproblemsolving.com/6/1/3/61354bd4e4dda19a09c20090538ac5654eedf7d5.png)
![\[2(a+b+c)\sum_{cyc}a\frac{\sqrt{bc(a+b+c)(-a+b+c)}}{b+c}\ge \frac{3(a+b)(b+c)(c+a)}{abc}\prod_{cyc}\frac{\sqrt{ bc(a+b+c)(-a+b+c)}}{b+c}\]](http://latex.artofproblemsolving.com/d/0/7/d070490741a9a604760fa24df753276b90649330.png)
![\[2\sum_{cyc}\frac{a}{b+c}\sqrt{bc(-a+b+c)}\ge 3\sqrt{\prod_{cyc}(-a+b+c)}\]](http://latex.artofproblemsolving.com/c/d/9/cd966eef6589de52b3cb3f53bd1d65fb95b4f662.png)
![\[\sum_{cyc}\frac{x+y}{x+y+2z}\sqrt{\frac{(x+z)(y+z)}{xy}}\ge 3 (leta=x+y,b=y+z,c=z+x)\]](http://latex.artofproblemsolving.com/b/5/1/b518198fe08d5cc4ae4dcd853e4944d94b57d62a.png)
![\[\sum_{cyc}\frac{(x+y)\sqrt{(x+z)(y+z)}-(x+y)\sqrt{xy}-2\sqrt{xy}z}{(x+y+2z)\sqrt{xy}}\ge 0\]](http://latex.artofproblemsolving.com/8/5/4/8546b0d94b444b413906973e2095a851e6d122ee.png)
![\[\sum_{cyc}\frac{(x+y)(x+y+z)z-2z(\sqrt{xy}+\sqrt{(x+z)(y+z)})\sqrt{xy}}{(x+y+2z)\sqrt{xy}(\sqrt{(x+z)(y+z)}+\sqrt{xy})}\ge 0\]](http://latex.artofproblemsolving.com/6/d/c/6dc7be35fa13a7b78056b50023158aa015da429d.png)
![\[\sum_{cyc}\frac{z(\sqrt{x(x+z)}-\sqrt{y(y+z)})^2}{(x+y+2z)\sqrt{xy}(\sqrt{(x+z)(y+z)}+\sqrt{xy})}\ge 0\]](http://latex.artofproblemsolving.com/b/5/d/b5db32f23d2c61d9d66d1f70eed59652443f4c8f.png)
,the inequality becomes
\ge 0$$](http://latex.artofproblemsolving.com/3/1/b/31bf5eda1d262b79495f11466fbf3998c1010bff.png)
(y-z)^2\ge 0$$](http://latex.artofproblemsolving.com/0/2/8/0286bbea436ac4bcf2cfcb6944dc8f9e33c8d0c6.png)
,we only need to prove that+x^2[y^2(x+z)-y^3+2y(x^2+xz+z^2)+xyz+2xz(x+z)](4y+x+z)\ge 0$$](http://latex.artofproblemsolving.com/7/7/2/7724dce3bf80ab7a3785bb83c92a99a79f5205ef.png)
.
.
, 
.
only need to prove when 
,which leads to 
.
.
.
when 
it becomes 
and when 
.
.
.
.
![\[{\frac {\sqrt {\Delta}}{{\it w_a}}}\geq {\frac {{3}^{3/4} \left( {\it r_a}+{\it
h_b}+{\it h_c} \right) }{2\,{\it h_c}+2\,{\it h_a}+{\it r_a}+{\it r_b}+2\,{
\it h_b}+{\it r_c}}}\]](http://latex.artofproblemsolving.com/2/1/e/21e506bd34e3829c9cc41a1f01bc47200b6724d7.png)
![\[\sqrt{\frac{l_a}{r_a}} \geq \frac{3(r+r_b))(r+r_c)}{s^2+r(5r+8R)}\]](http://latex.artofproblemsolving.com/1/c/0/1c0dd977d1654f2f3fe6b7df4994626680c10237.png)
![\[\left( r+{\it r_b} \right) \left( r+{\it r_c} \right) + \left( r+{\it
r_c} \right) \left( r+{\it r_a} \right) + \left( r+{\it r_a} \right)
\left( r+{\it r_b} \right) ={s}^{2}+r \left( 5\,r+8\,R \right) \]](http://latex.artofproblemsolving.com/e/c/0/ec0877439ba81a03b8794da9c5eed4815ad9793e.png)
![\[\sum m_bm_c \leqslant \frac{1}{4}(5p^2-20Rr+13r^2).\]](http://latex.artofproblemsolving.com/e/a/b/eabcf683deb61035e941777efd0f61fe00ad8db8.png)
We have![\[\sum m_a^2 = \frac{3}{4}(a^2+b^2+c^2) = \frac{1}{2}(3p^2-12Rr-3r^2)\]](http://latex.artofproblemsolving.com/f/9/c/f9c3d48af292ac674167d95de5ca208db95b7bcb.png)
and![\[\sum l_a^2 = \sum bc\left [ 1-\frac{a^2}{(b+c)^2} \right ] = \frac{p^6+3r^2p^4+(32R^2+40Rr+3r^2)r^2p^2+r^4(4R+r)^2}{(2Rr+p^2+r^2)^2},\]](http://latex.artofproblemsolving.com/5/9/c/59c2067066d2b21ef40911fbdc29f1068a9fe9af.png)
so![\[2 \sum m_bm_c - \sum m_a^2 \leqslant \frac{1}{2}(5p^2-20Rr+13r^2)-\frac{1}{2}(3p^2-12Rr-3r^2) = p^2-4Rr+8r^2.\]](http://latex.artofproblemsolving.com/5/f/3/5f3a0a35f9b71e52c58aeb00833194974719b96a.png)
Therefore we need to show that![\[\frac{p^6+3r^2p^4+(32R^2+40Rr+3r^2)r^2p^2+r^4(4R+r)^2}{(2Rr+p^2+r^2)^2} \geqslant p^2-4Rr+8r^2\]](http://latex.artofproblemsolving.com/a/3/a/a3a83a89317cd4e5f35fe3323862c347ec663569.png)
equivalent to![\[r^2\left [2(22R^2+6Rr-7r^2)p^2+r(16R^3-20Rr^2-7r^3)-7p^4 \right ] \geqslant 0.\]](http://latex.artofproblemsolving.com/d/1/8/d18a3ee25ac3e7ee40f68cd0f8900f2edd860126.png)
It suffices to prove that![\[2(22R^2+6Rr-7r^2)p^2+r(16R^3-20Rr^2-7r^3)-7p^4 \geqslant 0,\]](http://latex.artofproblemsolving.com/d/9/7/d97d2d8c8543934e911754713d54c9fa130e602f.png)
equivalent to![\[(4R^2+4Rr+3r^2-p^2)(4Rr+11p^2-4r^2)+(4p^2+r^2)(p^2+5r^2-16Rr)+36p^2r(R-2r) \geqslant 0.\]](http://latex.artofproblemsolving.com/c/d/4/cd4bb85c03421dbd8fddff15688ae317246d6683.png)
Which is clearly by Euler's and Gerretsen's Inequality.
inequalities. But in order to get there, we need a little bit of imagination ![\[\boxed{ \triangle ABC\ \wedge\ \forall\ P\in\mathcal{P}\implies \boxed{ \frac {PA}a + \frac{PB}b + \frac{PC}c \ge \frac {a + b + c}{PA + PB + PC} } }\]](http://latex.artofproblemsolving.com/1/5/6/15691c4c9c6d7c6a7ee25fe6ff80d22eb982510b.png)
![\[\left(\sum PA\right)^2\left(\sum \frac {PA^2}{a^2} + 2\sum \frac {PB\cdot PC}{b c}\right) \ge (a + b + c)^2\ .\]](http://latex.artofproblemsolving.com/8/d/2/8d2a8792f6b177a1a2fc6197639f65f1c6134d89.png)
![\[
\boxed{
\begin{array}{l}
\displaystyle \frac {PA\cdot PB}{ab} + \frac {PB\cdot PC}{bc} + \frac {PC\cdot PA}{ca} \ge 1 \\\\
\displaystyle \frac{PA^2}{a^2} + \frac{PB^2}{b^2} + \frac{PC^2}{c^2} \ge \frac {a^4 + b^4 + c^4}{a^2b^2 + b^2c^2 + c^2a^2}
\end{array}
}
\]](http://latex.artofproblemsolving.com/9/4/4/944c4277a48776de2965bee40d39ccfbdafafef6.png)
![\[
\begin{array}{lc}
& \displaystyle \left(\sum PA\right)^2\left(\frac{\sum a^4}{\sum b^2c^2} + 2\right) \ge (a + b + c)^2 \\\\
\iff & \displaystyle \left(\sum PA\right)^2 \left(\sum a^2\right)^2 \ge (a + b + c)^2 \sum b^2c^2 \\\\
\iff & \displaystyle \boxed{\sum PA \ge \frac {a + b + c}{a^2 + b^2 + c^2}\cdot\sqrt{a^2b^2 + b^2c^2 + c^2a^2}}
\end{array}
\]](http://latex.artofproblemsolving.com/c/5/2/c52f306497d69246da034947e5fa8fc3b26e8adf.png)
![\[
\boxed{
\begin{array}{ccccccc}
\max\{A, B, C\}\le 120^{\circ} & \implies & \displaystyle \sum PA & \ge & \displaystyle \sqrt{\frac {\sum a^2 + 4S\sqrt{3}}2} & \stackrel{(\ast)}{\ge} & \displaystyle \frac {\sum a}{\sum a^2}\cdot\sqrt{\sum b^2c^2} \\\\
A \ge 120^{\circ} & \implies & \displaystyle \sum PA & \ge & \displaystyle b + c & \stackrel{(\ast\ast)}{\ge} & \displaystyle \frac {\sum a}{\sum a^2}\cdot\sqrt{\sum b^2c^2}
\end{array}
}\ .
\]](http://latex.artofproblemsolving.com/e/b/5/eb544ff96249927d385e2f703fdefc70e075a9d7.png)
and 
in order to complete the proof within the above-mentioned restrictions. For example, one can show ![\[\frac {\sum a^2 + 4S\sqrt{3}}2 \ge s^2 + 9r^2\]](http://latex.artofproblemsolving.com/4/7/4/474859b70f236a47af45cd04b96f0798c0c9de39.png)
.
. And I spent a couple of months thinking about it until I managed to find a counterexample using GeoGebra (picture attached). Basically, the idea is to note that if we take 
(or 
as in my pic) then the inequality 
breaks immediately – in fact it only holds if 
.
,
.
,
is acute-angle and the other one when 
which naturally implies 
.![\[\begin{array}{lll}
a^2PA + b^2PB + c^2PC & \ge & (b^2 + c^2)PA + b^2PB + c^2PC \\\\
& = & b^2(PA + PB) + c^2(PA + PC) \\\\
& \ge & b^2\cdot c + c^2\cdot b \\\\
& = & bc(b + c).
\end{array}
\]](http://latex.artofproblemsolving.com/8/6/9/869fe9e584b31d2e6180a3472501da34736fc69f.png)
which follows from AM-GM and the fact that in any triangle it holds that 
(see ![\[bc(b + c)\ge 2bc\sqrt{bc} = 2\sqrt{(bc)^3} > 2\sqrt{11^3r^6} > 72r^3,\]](http://latex.artofproblemsolving.com/f/2/3/f2358a0ba2df2a0d76845f43f3bb73467ee1566d.png)
such that 
and 
the following inequality holds:![\[\begin{array}{c}(b + c)x + (c + a)y + (a + b)z \ge 2\sqrt{(xy + yz + zx) (ab + bc + ca)}.\end{array}\]](http://latex.artofproblemsolving.com/8/5/4/854d58d032b255a0aa826fe539fd8da98d2a942c.png)
and 
and taking into account Hayashi's inequality, i.e. 
we easily obtain the following
be real numbers such that ![\[\begin{array}{c} \displaystyle (y + z)\frac {PA}a + (z + x)\frac {PB}b + (x + y)\frac {PC}c \ge 2\sqrt{xy + yz + zx}.\end{array}\]](http://latex.artofproblemsolving.com/d/b/3/db3853438298976ae8a20870b69162bf1cc0adc4.png)
,
and 
in the previously derived statement we get:![\[\begin{array}{c} \displaystyle
2a^3\cdot \frac {PA}a + 2b^3\cdot \frac {PB}b + 2c^3\cdot \frac {PC}c \ge 2\sqrt{\sum (c^3 + a^3 - b^3)(a^3 + b^3 - c^3)} \\\\
\iff a^2PA + b^2PB + c^2PC \ge \sqrt{2(a^3b^3 + b^3c^3 + c^3a^3) - (a^6 + b^6 + c^6)} \\\\
\iff a^2PA + b^2PB + c^2PC \ge 2r\sqrt{9s^4 - 6s^2(6R^2 + 6Rr + r^2) + r(4R +r)^3}.
\end{array}\]](http://latex.artofproblemsolving.com/3/9/3/39340c4535487a7fb1c7429ec62af72efa19c54c.png)
which is known as ![\[
\begin{array}{lc}
& 2r\sqrt{9s^4 - 6s^2(6R^2 + 6Rr + r^2) + r(4R +r)^3} \ge 72r^3 \\\\
\iff & \displaystyle s^4 - \frac 23\cdot (6R^2 + 6Rr + r^2)s^2 + \frac {r(4R + r)^3}9 - 144r^4\ge 0 \\\\
\iff & \displaystyle \left(s^2 - \frac {6R^2 + 6Rr + r^2}3\right)^2 \ge \frac 49\cdot (9R^4 + 2R^3r + 324r^4) \\\\
\iff & \displaystyle s^2\ge 2R^2 + 2Rr + \frac {r^2}3 + \frac 23\sqrt{9R^4 + 2R^3r + 324r^4}.
\end{array}
\]](http://latex.artofproblemsolving.com/8/2/6/8265935a50aeab1b00236df02a40beea4d28b7aa.png)
and this will finally complete the proof. 
![\[\begin{array}{c} \displaystyle \left( \frac {PA}x + \frac {PB}y + \frac{PC}z \right)^2 \ge 2\left(\frac {ab}{xy} + \frac {bc}{yz} + \frac {ca}{zx}\right) - \left(\frac{a^2}{x^2} + \frac {b^2}{y^2} + \frac {c^2}{z^2}\right).\end{array}\]](http://latex.artofproblemsolving.com/9/a/8/9a81a9bf240459b5fa3b5fe69f5103d84d45e31f.png)
in acute or right-angled triangles:
, inradius 
, semiperimeter 
and medians of lengths 
, 
, 
, the following inequality holds:![\[\boxed{m_a + m_b + m_c \ge \sqrt{\frac {5s^2 + 12Rr + 3r^2}2}}\ .\]](http://latex.artofproblemsolving.com/3/5/4/35419a162729e890c0c83117e634a9c6d1cb4bd5.png)
![\[
\boxed{
\begin{array}{c}
\displaystyle \frac 1{a + b} + \frac 1{b + c} + \frac 1{c + a} = \frac {5s^2 + 4Rr + r^2}{2s (s^2 + 2Rr + r^2)} \\\\
\displaystyle \sum \frac {(m_a + m_b) (m_a + m_c)}{(a + b) (a + c)} = \frac {\big(\sum m_a\big)^2 - r^2}{s^2 + 2Rr + r^2} - \frac 14
\end{array}
}\ .
\]](http://latex.artofproblemsolving.com/5/d/0/5d032e5d40296659800109c3f7cea134f5ed8666.png)
![\[
\sum m_a \ge \frac {9s\sqrt{3} (s^2 + 2Rr + r^2)}{2(5s^2 + 4Rr + r^2)}
\]](http://latex.artofproblemsolving.com/7/7/7/7770aa72edf11fd0208db8592618959b14663863.png)
![\[
2(5s^2 + 12Rr + 3r^2)(5s^2 + 4Rr + r^2)^2 \ge 243 s^2 (s^2 + 2Rr + r^2)^2
\]](http://latex.artofproblemsolving.com/6/a/2/6a28f7eebc90a262da2df1ce1b3455dcfb63b865.png)
![\[X\cdot s^6 + Y\cdot s^4 + Z\cdot s^2 + W\ge 0.\]](http://latex.artofproblemsolving.com/3/7/0/370d50a54c6b424e80e77a98f101c2a1ffe5a811.png)
power to the 
by using Gerretsen's inequality i.e. 
and what remains to show will be somewhat easier to deal with. However, the computations are quite messy and I will refrain from writing them down here.![\[\sum m_a \ge \sqrt{\frac {5s^2 + 10Rr + 7r^2}2}\]](http://latex.artofproblemsolving.com/4/b/d/4bdf9fee3c67c0410501ce22d2232bff19bb4158.png)
is well-known.
![\[{\it w_a}\,{\it w_b}\,{\it w_c}+{\it h_a}\,{\it h_b}\,{\it h_c}\geq 2\, \left( {
\it m_b}+{\it m_c}-{\it m_a} \right) \left( {\it m_c}+{\it m_a}-{\it m_b}
\right) \left( {\it m_a}+{\it m_b}-{\it m_c} \right)\geq 2{\it h_a}\,{\it h_b}\,{\it h_c} \]](http://latex.artofproblemsolving.com/0/8/8/0881075d2d3d46653a5fa9a270d71c7eec28653b.png)
![\[ \left( {
\it m_b}+{\it m_c}-{\it m_a} \right) \left( {\it m_c}+{\it m_a}-{\it m_b}
\right) \left( {\it m_a}+{\it m_b}-{\it m_c} \right)\geq {\it h_a}\,{\it h_b}\,{\it h_c}\]](http://latex.artofproblemsolving.com/0/7/7/0773f4c0572d594434d8849d776b5af366b9e084.png)
<=>![\[{\it m_a}\,{\it m_b}\,{\it m_c}\geq r{s}^{2}\]](http://latex.artofproblemsolving.com/9/3/5/9352b2fa6ec4fa400eb1453b7031b227fe8a108a.png)
and easy![\[ \left( {\it m_b}+{\it m_c}-{\it m_a} \right) \left( {\it m_c}+{\it m_a}-{
\it m_b} \right) \left( {\it m_a}+{\it m_b}-{\it m_c} \right) \geq {\frac {9}
{8}}\,{\it h_a}\,{\it h_b}\,{\it h_c}-{\frac {27}{8}}\,{r}^{3}\]](http://latex.artofproblemsolving.com/c/c/b/ccb970d24616e79620ec8db910a34b1300040fe4.png)
,![\[\sqrt {{\frac {10\,x+9}{x+y}}}+\sqrt {{\frac {10\,y+9}{y+z}}}+\sqrt {{
\frac {10\,z+9}{z+x}}}\geq \frac{3}{2}\,\sqrt {38}\]](http://latex.artofproblemsolving.com/d/4/3/d434f896c910e66eb17e1fce636be1b1574e75c2.png)
,![\[\left( {\it \sum} \left( 1+6\,x+2\,y+5\,z \right) \right) ^{3}
\left( {\it \sum} \left( {\frac { \left( x+y \right) \left( 1+6\,x+2
\,y+5\,z \right) ^{3}}{10\,x+9}} \right) \right) ^{-1}\geq {\frac {171}{2
}}\]](http://latex.artofproblemsolving.com/0/9/f/09f8df563981b4a4c836b1b795aebbbc8d37e025.png)
,
