Are you sure this ineq true ?

Knowing that in any triangle $a+b+c \geq 3\sqrt{3} R$ (this was already posted - even recently - look for topic in "inequalities unsolved) it is enough to prove: $\sum \frac{a^2}{b+c-a} \geq a+b+c$ Consider casual substitution: $x=b+c-a$, $y=c+a-b$, $z=b+a-c$, so we have: $\sum \frac{(y+z)^2}{x} \geq 4(x+y+z)$ as inequality is homogenous we can add constraint $x+y+z=1$ so we're about to prove: $\frac{1}{3} \sum \frac{(1-x)^2}{x} \geq \frac{4}{3}$ But this follows from Jensen as function $f(x)=\frac{(1-x)^2}{x}$ is convex

Come on,guys.The inequality that Lagrangia posted is too easy.It follows immediately from Chebyshef, $a^2+b^2+c^2\geq 18Rr$ and from $\sum\frac{1}{b+c-a}\geq\frac{p\sqrt{3}}{r}$. The last inequality can be easy deduced from this inequality $xy+yz+zx\geq\sqrt{3xyz(x+y+z)}$ by putting $x=p-a,y=p-b,z=p-c$. By the way I have another two solutions for this one.

By the way,Megus you are totally wrong here. It is $a+b+c\leq 3\sqrt{3} R$

Aaaaa, by the way Megus it is too complicated and it isn't worth to do that substitutions since the inequality you supposed it is stronger follows imediately from Cauchy-Schwarz inequality.

You're right Cezar - sorry for this nonsenses above It was like I just remember I've seen this $3\sqrt{3} R$ I opened that topic and looked just at first post which firmly stands $\geq$ so I didn't read further ... thanks for pointing out

There is even a stronger one,but kind of ugly.

cezar lupu wrote: Come on,guys.The inequality that Lagrangia posted is too easy.It follows immediately from Chebyshef, $a^2+b^2+c^2\geq 18Rr$ and from $\sum\frac{1}{b+c-a}\geq\frac{p\sqrt{3}}{r}$. The last inequality can be easy deduced from this inequality $xy+yz+zx\geq\sqrt{3xyz(x+y+z)}$ by putting $x=p-a,y=p-b,z=p-c$. By the way I have another two solutions for this one. Oh Cezar , I infer nothing from two your ineqs

Cezar can you post the other two sols you found???? It looks interesting....

Hold on ! This problem is still unsolved

jarod wrote: Hold on ! This problem is still unsolved why?!

Why is this problem still unsolved? Jarod, I think there is a misunderstanding here. Can't you do all the calculations? I solved almost 85% of the problem.

Socrates I will post those two solutions very soon.

Megus, i think that $(a+b+c) \leq 3\sqrt{3}R$

That is true e.lopez. It is called Mitrinovici's inequality

Valentin Vornicu wrote: That is true e.lopez. It is called Mitrinovici's inequality i know.

e.lopes have you seen post number 5?

cezar lupu wrote: e.lopes have you seen post number 5? sorry

cezar lupu wrote: Come on,guys.The inequality that Lagrangia posted is too easy.It follows immediately from Chebyshef, $a^2+b^2+c^2\geq 18Rr$ and from $\sum\frac{1}{b+c-a}\geq\frac{p\sqrt{3}}{r}$. The last inequality can be easy deduced from this inequality $xy+yz+zx\geq\sqrt{3xyz(x+y+z)}$ by putting $x=p-a,y=p-b,z=p-c$. By the way I have another two solutions for this one. can you detail it? and i don't think $\sum\frac{1}{b+c-a}\geq\frac{p\sqrt{3}}{r}$ is right...

Yes, you are right. Actually, it is $\sum\frac{1}{b+c-a}\geq\frac{\sqrt{3}}{2r}.$

The solution i gave can be found in my book of inequalities. It is problem GD.56

Lagrangia wrote: Prove that in any triangle $ABC$ the following inequality holds: $\frac{a^{2}}{b+c-a}+\frac{b^{2}}{a+c-b}+\frac{c^{2}}{a+b-c}\geq 3\sqrt{3}R$. The above inequality can be refined as follows: \[\sum\frac{a^{2}}{b+c-a}\geq\frac{R}{2r}(a+b+c)\geq\frac{1}{4}\sum\frac{(b+c)^{2}}{b+c-a}\geq 3\sqrt{3}R, \] where $a,b,c$ are the sides of triangle $ABC$, $R$ is radius of the circumcircle and $r$ is radius of the incircle. The second inequality see: http://www.mathlinks.ro/Forum/viewtopic.php?t=63848

$ a+b+c\geq 3\sqrt{3}R $ is not right

Lagrangia wrote: Prove that in any triangle $ABC$ the following inequality holds: \[ \frac{a^{2}}{b+c-a}+\frac{b^{2}}{a+c-b}+\frac{c^{2}}{a+b-c}\geq 3\sqrt{3}R. \] Laurentiu Panaitopol Let $a+b-c=z$, $a+c-b=y$, $b+c-a=x$, $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$. Hence, $\sum_{cyc}\frac{a^2}{b+c-a}\geq3\sqrt{3}R\Leftrightarrow\sum_{cyc}\frac{a^2}{b+c-a}\geq\frac{3\sqrt{3}abc}{4S}\Leftrightarrow$ $\Leftrightarrow\sum_{cyc}\frac{(y+z)^2}{4x}\geq\frac{3\sqrt3(x+y)(x+z)(y+z)}{8\sqrt{xyz(x+y+z)}}\Leftrightarrow$ $\Leftrightarrow2\sum_{cyc}xy(x+y)^2\geq3(x+y)(x+z)(y+z)\sqrt{\frac{3xyz}{x+y+z}}\Leftrightarrow$ $\Leftrightarrow2(27u^2v^2-15uw^3)\geq3(9uv^2-w^3)\sqrt{\frac{w^3}{u}}\Leftrightarrow f(v^2)\geq0$, where $f$ is a linear function. Hence, by $uvw$ it remains to check only one case: $y=z=1$, which gives $2(2x^3+2x+2+4x^2+2)\geq3\cdot2(x+1)^2\sqrt{\frac{3x}{x+2}}$, which is $(x-1)^2(4x^5+32x^4+89x^3+118x^2+85x+32)\geq0$, which is obvious and very very not strong. Indeed, even the Ji Chen's following stronger inequality is very easy. Ji Chen wrote: The above inequality can be refined as follows: \[\sum\frac{a^{2}}{b+c-a}\geq\frac{R}{2r}(a+b+c)\] It's equivalent to $\sum_{cyc}(x^3y+x^3z+2x^2y^2-4x^2yz)\geq0$.

My proof for Ji Chen's following inequality: \sum\frac{a^2}{b+c-a}\geq\frac{R}(2r}(a+b+c) (1) Using the usual notations ,(1) is equivalent to: \a^2r_a+b^2r_b+c^2r_c\geq\2s^2R (2) Applying Chebyshef's inequality we get: \a^2r_a+b^2r_b+c^2r_c\geq\frac{1}{3}(a^2+b^2+c^2)(r_a+r_b+r_c) (3) It suffices to show that: \(a^2+b^2+c^2)(r_a+r_b+r_c\geq\6s^2R (4) Are well-known relationships a^2+b^2+c^2=2(s^2-r^2-4Rr and r_a+r_b+r_c=r+4R,so (4) is rewritten as: \(s^2-r^2-4Rr)(r+4R)\geq\ 3s^2R (5) After calculations we obtain: \ s^2\ geq\ frac{r(r+4R)^2}{R+r} (6) which is well-known.

Lagrangia wrote: Prove that in any triangle $ABC$ the following inequality holds: \[ \frac{a^{2}}{b+c-a}+\frac{b^{2}}{a+c-b}+\frac{c^{2}}{a+b-c}\geq 3\sqrt{3}R. \] Laurentiu Panaitopol Bosnia and Herzegovina Team Selection Test 1999

Lagrangia wrote: Prove that in any triangle $ABC$ the following inequality holds: \[ \frac{a^{2}}{b+c-a}+\frac{b^{2}}{a+c-b}+\frac{c^{2}}{a+b-c}\geq 3\sqrt{3}R. \] Laurentiu Panaitopol this hold: In triangle,prove that \[{\frac {{a}^{2}}{b+c-a}}+{\frac {{b}^{2}}{a+c-b}}+{\frac {{c}^{2}}{a+b -c}}\geq 3\,R \left( \cot \left( A \right) +\cot \left( B \right) +\cot \left( C \right) \right) \]

Lagrangia wrote: Prove that in any triangle $ABC$ the following inequality holds: \[ \frac{a^{2}}{b+c-a}+\frac{b^{2}}{a+c-b}+\frac{c^{2}}{a+b-c}\geq 3\sqrt{3}R. \] Laurentiu Panaitopol hard is this: \[ \sum{\frac{a^{2}}{b+c-a}\cos^5{\frac{B-C}{2}}}\geq 3\sqrt{3}R. \]

xzlbq wrote: In triangle,prove that \[{\frac {{a}^{2}}{b+c-a}}+{\frac {{b}^{2}}{a+c-b}}+{\frac {{c}^{2}}{a+b -c}}\geq 3\,R \left( \cot \left( A \right) +\cot \left( B \right) +\cot \left( C \right) \right) \] Note that \[\sum \frac{a^2}{b+c-a} = \frac{2p(R-r)}{r},\]and \[\sum \cot A = \frac{p^2-4Rr-r^2}{2pr}.\]Therefore we need to prove \[12R^2r+Rp^2+3Rr^2-4p^2r \geqslant 0,\]equivalent to \[(4Rr+p^2+3r^2)(R-2r)+2r(4R^2+4Rr-p^2+3r^2) \geqslant 0.\]Which is true.

xzlbq wrote: this hold: In triangle,prove that \[{\frac {{a}^{2}}{b+c-a}}+{\frac {{b}^{2}}{a+c-b}}+{\frac {{c}^{2}}{a+b -c}}\geq 3\,R \left( \cot \left( A \right) +\cot \left( B \right) +\cot \left( C \right) \right) \] In triangle,Let $Q=R \left( {\frac {a}{{\it h_a}-r}}+{\frac {b}{{\it h_b}-r}}+{\frac {c}{{ \it h_c}-r}} \right) $ then prove that \[{\frac {{a}^{2}}{b+c-a}}+{\frac {{b}^{2}}{a+c-b}}+{\frac {{c}^{2}}{a+b -c}}\geq Q\geq 3\,R \left( \cot \left( A \right) +\cot \left( B \right) +\cot \left( C \right) \right) \]

xzlbq wrote: In triangle,Let $Q=R \left( {\frac {a}{{\it h_a}-r}}+{\frac {b}{{\it h_b}-r}}+{\frac {c}{{ \it h_c}-r}} \right) $ then prove that \[{\frac {{a}^{2}}{b+c-a}}+{\frac {{b}^{2}}{a+c-b}}+{\frac {{c}^{2}}{a+b -c}}\geq Q \] The Ravi's substitution gives $$(x+y+z)\sum_{sym}(3x^4y^2-x^4yz+3x^3y^3-2x^3y^2z-3x^2y^2z^2)\geq0,$$which is Muirhead.

Right side is: In triangle,Let $Q=R \left( {\frac {a}{{\it h_a}-r}}+{\frac {b}{{\it h_b}-r}}+{\frac {c}{{ \it h_c}-r}} \right) $ then prove that \[Q\geq 3\,R \left( \cot \left( A \right) +\cot \left( B \right) +\cot \left( C \right) \right) \] <=> \[{\frac {{a}^{2} \left( 2\,a-b-c \right) }{b+c}}+{\frac {{b}^{2} \left( 2\,b-c-a \right) }{c+a}}+{\frac {{c}^{2} \left( 2\,c-a-b \right) }{a+b}}\geq 0\] How to do?

xzlbq wrote: \[{\frac {{a}^{2} \left( 2\,a-b-c \right) }{b+c}}+{\frac {{b}^{2} \left( 2\,b-c-a \right) }{c+a}}+{\frac {{c}^{2} \left( 2\,c-a-b \right) }{a+b}}\geq 0\] $$\sum_{cyc}\frac{a^2(2a-b-c)}{b+c}=\sum_{cyc}(a-b)\left(\frac{a^2}{b+c}-\frac{b^2}{a+c}\right)=\sum_{cyc}\frac{(a-b)^2(a^2+ab+b^2+ac+bc)}{(a+c)(b+c)}\geq0.$$

xzlbq wrote: Right side is: In triangle,Let $Q=R \left( {\frac {a}{{\it h_a}-r}}+{\frac {b}{{\it h_b}-r}}+{\frac {c}{{ \it h_c}-r}} \right) $ then prove that \[Q\geq 3\,R \left( \cot \left( A \right) +\cot \left( B \right) +\cot \left( C \right) \right) \] <=> \[{\frac {{a}^{2} \left( 2\,a-b-c \right) }{b+c}}+{\frac {{b}^{2} \left( 2\,b-c-a \right) }{c+a}}+{\frac {{c}^{2} \left( 2\,c-a-b \right) }{a+b}}\geq 0\] How to do? \[{\frac {{a}^{2} \left( 2\,a-b-c \right) }{b+c}}+{\frac {{b}^{2} \left( 2\,b-c-a \right) }{c+a}}+{\frac {{c}^{2} \left( 2\,c-a-b \right) }{a+b}}\geq 0\] <=> \[{\frac {{a}^{2}}{b+c}}+{\frac {{b}^{2}}{c+a}}+{\frac {{c}^{2}}{a+b}}\geq \frac{3} {4}\,{\frac {{a}^{2}+{b}^{2}+{c}^{2}}{s}}\]But prove this stronger: \[{\frac {{a}^{2} \left( a+b+c \right) }{ \left( b+c \right) \left( {b} ^{2}+{c}^{2}+{a}^{2} \right) }}\geq \frac{3}{2}\,{\frac {{a}^{2} \left( a \left( b +c \right) - \left( b-c \right) ^{2} \right) }{{\it \sum} \left( {a}^{ 2} \left( a \left( b+c \right) - \left( b-c \right) ^{2} \right) \right) }}\]

xzlbq wrote: xzlbq wrote: Right side is: In triangle,Let $Q=R \left( {\frac {a}{{\it h_a}-r}}+{\frac {b}{{\it h_b}-r}}+{\frac {c}{{ \it h_c}-r}} \right) $ then prove that \[Q\geq 3\,R \left( \cot \left( A \right) +\cot \left( B \right) +\cot \left( C \right) \right) \] <=> \[{\frac {{a}^{2} \left( 2\,a-b-c \right) }{b+c}}+{\frac {{b}^{2} \left( 2\,b-c-a \right) }{c+a}}+{\frac {{c}^{2} \left( 2\,c-a-b \right) }{a+b}}\geq 0\] How to do? \[{\frac {{a}^{2} \left( 2\,a-b-c \right) }{b+c}}+{\frac {{b}^{2} \left( 2\,b-c-a \right) }{c+a}}+{\frac {{c}^{2} \left( 2\,c-a-b \right) }{a+b}}\geq 0\] <=> \[{\frac {{a}^{2}}{b+c}}+{\frac {{b}^{2}}{c+a}}+{\frac {{c}^{2}}{a+b}}\geq \frac{3} {4}\,{\frac {{a}^{2}+{b}^{2}+{c}^{2}}{s}}\]But prove this stronger: \[{\frac {{a}^{2} \left( a+b+c \right) }{ \left( b+c \right) \left( {b} ^{2}+{c}^{2}+{a}^{2} \right) }}\geq \frac{3}{2}\,{\frac {{a}^{2} \left( a \left( b +c \right) - \left( b-c \right) ^{2} \right) }{{\it \sum} \left( {a}^{ 2} \left( a \left( b+c \right) - \left( b-c \right) ^{2} \right) \right) }}\] Please by C-S or A-G do this: \[{\frac {{a}^{2}}{b+c}}+{\frac {{b}^{2}}{c+a}}+{\frac {{c}^{2}}{a+b}}\geq \frac{3} {4}\,{\frac {{a}^{2}+{b}^{2}+{c}^{2}}{s}}\]