Let $ cotA=x,cotB=y,cotC=z $ Then this problem becomes: $ a,b,c \ge 0 $ satisfy $ ab+bc+ca=1 $, prove that: $ \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \ge \frac{5}{2} $ which is from China TST 2005.

BSJL wrote: Let $ cotA=x,cotB=y,cotC=z $ Then this problem becomes: $ a,b,c \ge 0 $ satisfy $ ab+bc+ca=1 $, prove that: $ \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \ge \frac{5}{2} $ which is from China TST 2005. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=449840 2003 China Mathematical Olympiad team test questions 2008 China Jiangxi mathematics competition problems

v_Enhance wrote: In a non-obtuse triangle $ABC$, prove that \[ \frac{\sin A \sin B}{\sin C} + \frac{\sin B \sin C}{\sin A} + \frac{\sin C \sin A}{ \sin B} \ge \frac 52. \]Proposed by Ryan Alweiss http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=537284

sqing wrote: v_Enhance wrote: In a non-obtuse triangle $ABC$, prove that \[ \frac{\sin A \sin B}{\sin C} + \frac{\sin B \sin C}{\sin A} + \frac{\sin C \sin A}{ \sin B} \ge \frac 52. \]Proposed by Ryan Alweiss http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=537284 \[ \frac{\sin A \sin B}{\sin C} + \frac{\sin B \sin C}{\sin A} + \frac{\sin C \sin A}{ \sin B} \ge \frac 52+(12\sqrt{3}-20)\cos{A}\cos{B}\cos{C}. \] actute triangle

I shall use the method of Lagrange Multipliers to show that if for some nonnegative reals $ a, b, c $ we have $ ab + bc + ca = 1 $ then $ \frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a} \geq \frac{5}{2} $. Define $ f(a, b, c) = -\frac{1}{a + b} - \frac{1}{b + c} - \frac{1}{c + a} $ and $ g(a, b, c) = ab + bc + ca $. Note that $ f $ and $ g $ and all of their partial derivatives are continuous. Moreover, note that $ \nabla {g} \neq 0 $ at all points since at most one of $ a, b, c $ can equal $ 0 $. Now, define the open set: \[ U =\left\{ (a,b,c)\mid 0 < a,b,c < 1000 \right\}. \]Its closure is: \[ \overline U =\left\{ (a,b,c)\mid 0 \leq a,b,c \leq 1000 \right\}. \]Therefore the constraint set: \[ \overline S =\left\{\mathbf{x}\in \overline U : g( \mathbf{x}) = 1\right\} \]Is compact, and so $ f $ obtains a global maximum over this set. Now if $ \mathbf{x} $ lies on the boundary, one of $ a, b, c $ must clearly be equal to $ 0 $ (if one of $ a, b, c $ is equal to $ 1000 $ the claim is trivial). In this case, it suffices to show that if $ ab = 1 $ then $ \frac{1}{a} + \frac{1}{b} + \frac{1}{a + b} \geq \frac{5}{2} $ but this is trivial. Now, we assume $ \mathbf{x} \in U $ and consider a local maximum. Then we must have $ \left<\frac{1}{(a + b)^2} + \frac{1}{(a + c)^2}, \frac{1}{(b + c)^2} + \frac{1}{(b + a)^2}, \frac{1}{(c + a)^2} + \frac{1}{(c + b)^2}\right> $ equal to $ \lambda\left<b + c, c + a, a + b\right> $ for some real $ \lambda $. Now let $ x = b + c, y = c + a, z = a + b $. We have $ \left<\frac{1}{z^2} + \frac{1}{y^2}, \frac{1}{x^2} + \frac{1}{z^2}, \frac{1}{y^2} + \frac{1}{x^2}\right> $ equal to $ \lambda\left<x, y, z\right> $. Since $ \frac{1}{z^2} + \frac{1}{y^2} = \lambda{x} $ and $ \frac{1}{x^2} + \frac{1}{z^2} = \lambda{y} $, subtracting these two equations yields that either $ x = y $ or that $ \frac{x + y}{x^2y^2} = \lambda $. For now, assume that $ x \neq y \neq z \neq x $. Then $ \frac{x + y}{x^2y^2} = \frac{x + z}{x^2z^2} $ which, upon expanding and dividing both sides by $ y - z $, implies that $ xy + yz + zx = 0 $. But because $ ab + bc + ca = 1 $ we know that $ 2xy + 2yz + 2zx - x^2 - y^2 - z^2 = 4 $. Therefore we must have $ - x^2 - y^2 - z^2 = 4 $, absurd. Therefore at least two of $ x, y, z $ must be equal, which implies that at least two of $ a, b, c $ are equal. Assume WLOG that $ a = c $. It suffices to show that if $ a^2 + 2ab = 1 $ then $ \frac{1}{2a} + \frac{2}{a + b} \geq \frac{5}{2} $. Solving for $ a $ in terms of $ b $, substituting and clearing denominators it suffices to show that if $ 0 \leq a \leq 1 $ then $ 5a^3 - 9a^2 + 5a - 1 = (a - 1)(5a^2 - 4a + 1) \leq 0 $. So it suffices to show that $ 5a^2 - 4a + 1 \geq 0 $. But since $ 5a^2 - 4a + 1 = 5\left(a - \frac{2}{5}\right)^2 + \frac{1}{5} \geq \frac{1}{5} > 0 $, we are done. Phew!

BSJL wrote: Let $ cotA=x,cotB=y,cotC=z $ Then this problem becomes: $ a,b,c \ge 0 $ satisfy $ ab+bc+ca=1 $, prove that: $ \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \ge \frac{5}{2} $ which is from China TST 2005. I couldn't find it in the contests section

utkarshgupta wrote: BSJL wrote: Let $ cotA=x,cotB=y,cotC=z $ Then this problem becomes: $ a,b,c \ge 0 $ satisfy $ ab+bc+ca=1 $, prove that: $ \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \ge \frac{5}{2} $ which is from China TST 2005. I couldn't find it in the contests section http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=449840 2003 China Mathematical Olympiad team test questions 2008 China Jiangxi mathematics competition problems \[\left(\frac{\sin A\sin B}{\sin C}\right)^{2}+\left(\frac{\sin B\sin C}{\sin A}\right)^{2}+\left(\frac{\sin C\sin A}{\sin B}\right)^{2}\geq\frac94,\] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=142583&p=3651217#p3651217

After transforming to the inequality $\sum \frac{1}{a+b}\ge \frac{5}{2}$, there is a short solution. The inequality expands to \[2(\sum a^2 +3\sum ab)\sqrt{ab+bc+ca}\ge 5(\sum_{sym} a^2b +2abc).\] In terms of $uvw$ this is \[2\sqrt{3} (9u^2+3v^2)v\ge 5(9uv^2-w^3)\] which is linear in $w^3$ when fixing $u,v$. Hence it suffices to check when $a=b=1$ and when $c=0$. The first case goes $2(c^2+6c+5)\sqrt{2c+1}\ge 10(c+1)^2$. We can factor out $2(c+1)$ and we want $(c+5)\sqrt{2c+1}\ge 5(c+1)$. After squaring and simplifying this is $c(c^2-2c+5)\ge 0$ which is true with equality $c=0$. The second case goes $2(a^2+b^2+3ab)\sqrt{ab}\ge 5(a^2b+ab^2)$. WLOG $b=1$ and squaring and simplifying, we get $4a^5-a^4-6a^3-a^2+4a\ge 0$, which factors as $a(a-1)^2(4a^2+7a+4)\ge 0$, which is clearly true. If $a=0$ then clearly $ab+bc+ca\neq 1$, so $a=1$. Hence the inequality is true, with equality case $(1,1,0)$ and permutations.

This problem looks harmless, but it's anything but that... Anyways, here is a solution that doesn't use the cotangent substitution. It's somewhat bashy, but there's definitely some ingenuity in here as well.

BSJL wrote: Let $ cotA=x,cotB=y,cotC=z $ Then this problem becomes: $ a,b,c \ge 0 $ satisfy $ ab+bc+ca=1 $, prove that: $ \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \ge \frac{5}{2} $ which is from China TST 2005. The following one is also true $x,y,z \ge0 $ and $xy+yz+zx=1$, prove that \[\sum_{cyc} \frac{1}{x+y} \geqslant \sqrt{\frac{25}{4}+\frac92 \cdot \frac{xyz}{x+y+z}}\]

I created this problem by using the China TST 2005. It's probably not the best problem ever.

bedwarmer wrote: The following one is also true $x,y,z \ge0 $ and $xy+yz+zx=1$, prove that \[\sum_{cyc} \frac{1}{x+y} \geqslant \sqrt{\frac{25}{4}+\frac92 \cdot \frac{xyz}{x+y+z}}\] Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where and $xyz=w^3$. Hence, $\sum_{cyc} \frac{1}{x+y} \geqslant \sqrt{\frac{25}{4}+\frac92 \cdot \frac{xyz}{x+y+z}}\Leftrightarrow\frac{(9u^2+3v^2)\sqrt{3v^2}}{9uv^2-w^3}\geq\sqrt{\frac{25}{4}+\frac{w^3}{2uv^2}}\Leftrightarrow f(w^3)\geq0$, where $f(w^3)=2\sqrt{3u}v^2(9u^2+3v^2)-(9uv^2-w^3)\sqrt{25uv^2+2w^3}$. $f'(w^3)=\sqrt{25uv^2+2w^3}-\frac{9uv^2-w^3}{\sqrt{25uv^2+2w^3}}=\frac{16uv^2+3w^3}{\sqrt{25uv^2+2w^3}}>0$. Hence, by $uvw$ it remains to prove our inequality in two cases: 1. $z=0$, $y=\frac{1}{x}$, which gives something obvious; 2. $y=x$ and $z=\frac{1-x^2}{2x}$, where $0<x\leq1$, which gives $(3x^2-1)^2(1+3x^2-2x^4)\geq0$, which is obvious.

Like everyone else substitute $\cot A=x$ and similarly for $y, z$; we're trying to prove that for nonnegative reals $x, y, z$ satisfying $xy+yz+zx=1$ the inequality $\frac{1}{x+y} + \frac{1}{y+z}+\frac{1}{z+x}\geq \frac{5}{2}$ holds. Multiply both sides by $2(x+y)(y+z)(z+x)$, expand, and rearrange to obtain \[6(xy+yz+zx)+2(x^2+y^2+z^2)-10xyz-5\sum_{\text{cyc}}x^2(y+z)\geq 0.\]Using the condition we have $x^2(y+z)=x^2\cdot \frac{(1-yz)}{x}=x-xyz$ and $x^2+y^2+z^2=(x+y+z)^2-2$, so the above turns into \[2(x+y+z)^2-5(x+y+z)+2+5xyz\geq 0\implies (2(x+y+z)-1)(x+y+z-2)+5xyz\geq 0.\] Set $x+y+z=a$. If $a\geq 2$ then the inequality obviously holds. Assume $a<2$. Then from Schur we have \[(x+y+z)^3+9xyz\geq 4(x+y+z)(xy+yz+zx)\implies 9xyz\geq 4a-a^3\]so it suffices to prove that \[2a^2-5a+2+\frac{5}{9}(4a-a^3)\geq 0\implies (2-a)\left(5\left(a-\frac{4}{5}\right)^2+\frac{29}{5}\right)\geq 0,\]which is obviously true. We are done.

How is this even A1? Anyways, Can someone clear my doubt please? Is this true for any positive reals $x,y,z$? $\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\geq\frac{5}{4\sqrt{(x+y+z)xyz}}$?

I guess you can maybe frame a stronger inequality, though I don't know how.

Math-wiz wrote: How is this even A1? Anyways, Can someone clear my doubt please? Is this true for any positive reals $x,y,z$? $\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\geq\frac{5}{4\sqrt{(x+y+z)xyz}}$? Try $x\rightarrow0^+$.

Oops! Sorry, arqady

Solved with cj13609517288 First note that $\sin C = \sin(A+B)$. From this, we can make the substitution $\cot A = x$ (similarly with $y,z$), and simplifying yields the new problem $\frac{1}{x+y} + \frac{1}{y+z}+\frac{1}{z+x}\geq \frac{5}{2}$ with $xy+yz+zx=1$. Multiply and rearrange the inequality to get $$2x^2+2y^2+2z^2+6-10xyz \geq x^2y+z+xy^2+z+xyz^2$$This becomes $(2(x+y+z)-1)(x+y+z-2)+5xyz\geq 0$. by a factorization. By Schur, $$(x+y+z)^3+9xyz\geq 4(x+y+z)(xy+yz+zx)$$Set $n=x+y+z$. The problem reduces to showing $-\frac{5}{9}n^3+2n^2-\frac{25}{9}n+2\geq0$, which is clear.

Here is a solution without the cotangent substitution: The original inequality is equivalent to $$\frac{1}{a^2} +\frac{1}{b^2} +\frac{1}{c^2} \geq \frac{5}{4}(\triangle)^{-1}$$Do the Ravi-like substitution $a^2=x+y$ etc. for $x,y,z \in \mathbb{R}_{\geq 0}$ to get rid of the non-obtuse condition. $$ \triangle= \frac{1}{4}\sqrt{2\sum{a^2b^2} -\sum{a^4}}= \frac{1}{2}\sqrt{xy+yz+zx} $$So the inequality is equivalent to $$ (\sum x^2 + 3\sum xy)\sqrt{\sum xy}\geq \frac{5}{2}(x+y)(y+z)(z+x) $$Now graph and consider the intersection points of the cubic $f(p)=(p-x)(p-y)(p-z)-c$ with the x-axis. As $c$ gets bigger, the two smaller roots of $f$ approach each other. So for the largest possible postive real $c$ such that $f(p)$ has 3 positive roots $x',y',z'$ must satisfy the condition that at least two of them are equal. Going from $(x,y,z) \rightarrow (x',y',z')$ does not change the lhs of the inequality and makes the rhs bigger (expanding $f(p)$ or using vieta's theorems) , so we just need to show the inequality for the triplets of the form $(x,x,y)$ which is obvious after squaring and expanding.