I think you mean positive reals. This is very easy, I can't believe it was proposed in a recent olympiad.. Letting $ a = \frac xy,b = \frac yz, c = \frac zx$ implies \[ LHS = \sum_{cyc}\frac {x^2}{z^2 + xy}\ge\frac {(x^2 + y^2 + z^2)^2}{x^2y^2 + y^2z^2 + z^2x^2 + x^3y + y^3z + z^3x} \] Now it remains to prove that \[ 2(x^2 + y^2 + z^2)^2\ge 3\sum_{cyc}x^2y^2 + 3\sum_{cyc}x^3y \] Which follows by adding the two inequalities \[ x^4 + y^4 + z^4\ge x^3y + y^3z + z^3x \\ \\ \sum_{cyc}(x^4 + x^2y^2)\ge \sum_{cyc}2x^3y \] I bet all the contestants solved this one. EDIT: Here is a related topic: http://www.mathlinks.ro/viewtopic.php?search_id=1111061624&t=165948

the reason u r getting the solution easily is u have taken the LHS wrong... it should be $ \sum \frac {xz}{x^{2} + yz}$ P.S. laso how did u get ur first inequality that is nayel wrote: $ LHS = \sum_{cyc}\frac {x^2}{z^2 + xy}\ge\frac {(x^2 + y^2 + z^2)^2}{x^2y^2 + y^2z^2 + z^2x^2 + x^3y + y^3z + z^3x}$

pardesi wrote: the reason u r getting the solution easily is u have taken the LHS wrong... it should be $ \sum \frac {xz}{x^{2} + yz}$ No,nayel's solution is right. nayel wrote: $ LHS = \sum_{cyc}\frac {x^2}{z^2 + xy}\ge\frac {(x^2 + y^2 + z^2)^2}{x^2y^2 + y^2z^2 + z^2x^2 + x^3y + y^3z + z^3x}$ He used Cauchy-Schwarz ineqaulity

sorry i had copied the inequality wrong sorry nayel

and how is that from cauchy

pardesi wrote: and how is that from cauchy $ \sum\frac{x^2}{z^2+xy}=\sum\frac{x^4}{z^2x^2+x^3y}=\sum\frac{(x^2)^2}{z^2x^2+x^3y}$. P.S: Official solution is much nicer and easier! I hope that some of you will find it.

isn't it strange if we apply the engel form to $ \sum \frac{x^{2}}{z^{2}+yx} \geq \frac{(\sum x)^{2}}{\sum x^{2} + \sum xy}$ we don't get the desired result...we get a weak bound. u have to multiply by $ x^{2}$

pardesi wrote: isn't it strange if we apply the engel form to $ \sum \frac {x^{2}}{z^{2} + yx} \geq \frac {(\sum x)^{2}}{\sum x^{2} + \sum xy}$ we don't get the desired result...we get a weak bound. u have to multiply by $ x^{2}$ Yes,that is even mystical. That is the reason why,i didn't get 7 points for this problem, but anyway i've received gold medal

congrats anyways

pardesi wrote: isn't it strange if we apply the engel form to $ \sum \frac {x^{2}}{z^{2} + yx} \geq \frac {(\sum x)^{2}}{\sum x^{2} + \sum xy}$ we don't get the desired result...we get a weak bound. u have to multiply by $ x^{2}$ Multiplying by $ x^2$ isn't necessary. It's enough to multiply by $ x$. (But $ x^2$ makes things a bit more easier) \[ \sum_{cyc}\frac {x^3}{z^2x + x^2y}\ge\frac {(x^{\frac 32} + y^{\frac 32} + z^{\frac 32})^2}{2(x^2y + y^2z + z^2x)} \] Then it remains to show that \[ (x^{\frac 32} + y^{\frac 32} + z^{\frac 32})^2\ge 3(x^2y + y^2z + z^2x) \] Which follows from AM-GM \[ \sum_{cyc}(x^3 + x^{\frac 32}y^{\frac 32} + x^{\frac 32}y^{\frac 32})\ge \sum_{cyc}3x^2y \] PS. Congrats to Erken.

no it's not multiplying by $ x$ or $ x^{2}$...the thing is u can't put engel form directly without multiplying and dividing out the same...and get the answer

pardesi wrote: ...the thing is u can't put engel form directly without multiplying and dividing out the same...and get the answer That's what makes the problem a bit more standard. Otherwise it would be trivial right?

see my point was not about triviality...but the fact that until and unless u multiply it by $ x$ or $ x^{2}$ or so u don't get the solution is n't that amazing

Yes you're right that is interesting... but it's also a well known strategy.

OFFICIAL SOLUTION: LEMMA: Suppose that $ x_{1} > = x_{2} > = x_{3},y_{1} < = y_{2} < = y_{3}$, and $ (y_{1}^{'};y_{2}^{'};y_{3}^{'})$ is a permutation of $ (y_{1};y_{2};y_{3})$.Then $ x_{1}y_{1}^{'} + x_{2}y_{2}^{'} + x_{3}y_{3}^{'} > = x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3}$. Proof: Since $ y_{1} < = y_{2} < = y_{3}$, we get $ y_{1}^{'} > = y_{1}$ and $ y_{1}^{'} + y_{2}^{'} > = y_{1} + y_{2}$;moreover, $ y_{1}^{'} + y_{2}^{'} + y_{3}^{'} = y_{1} + y_{2} + y_{3}$. $ x_{1}y_{1}^{'} + x_{2}y_{2}^{'} + x_{3}y_{3}^{'} = (x_{1} - x_{2})y_{1}^{'} + (x_{2} - x_{3})(y_{1}^{'} + y_{2}^{'}) + x_{3}(y_{1}^{'} + y_{2}^{'} + y_{3}^{'}) > = (x_{1} - x_{2})y_{1} + (x_{2} - x_{3})(y_{1} + y_{2}) + x_{3}(y_{1} + y_{2} + y_{3}) = x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3}$, Q.E.D. Denote by $ S$ the left-hand part of the desired inequality.Since $ S$ is invariant under the cyclical permutation of variables, we can assume $ a < = b < = c$ or $ a > = b > = c$. In both cases, by LEMMA we get: $ S = \frac {1}{(a + b)b} + \frac {1}{(b + c)c} + \frac {1}{(c + a)a} > = \frac {1}{(a + b)c} + \frac {1}{(b + c)a} + \frac {1}{(c + a)b} = T$. Hence, $ 2S > = S + T = (\frac {1}{(a + b)c} + \frac {1}{(a + b)b}) + (\frac {1}{(b + c)a} + \frac {1}{(b + c)c}) + (\frac {1}{(c + a)b} + \frac {1}{(c + a)a}) = \frac {b + c}{(a + b)bc} + \frac {c + a}{(b + c)ca} + \frac {a + b}{(c + a)ab} > = 3(\frac {1}{abc})^{\frac {1}{3}} = 3$, Q.E.D.

$ \sum\frac{x^2}{y^2+xz}\geq\sum\frac{2x^2}{2y^2+x^2+z^2}$ from A.G.M and now applying C.S. in Engel form works $ \sum\frac{2x^2}{2y^2+x^2+z^2}\geq\frac{2(x^2+y^2+z^2)^2}{x^4+y^4+z^4+3x^2y^2+3y^2z^2+3x^2z^2}\geq\frac{2.3}{4}$

This is harder. Let $ a, b, c$ be positive numbers for which $ abc = 1$. Prove that $ \sum \frac {1}{b(5a +b)} \ge \frac {1}{2}$.

Vasc wrote: This is harder. Let $ a, b, c$ be positive numbers for which $ abc = 1$. Prove that $ \sum \frac {1}{b(5a + b)} \ge \frac {1}{2}$. Put $ a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$ where $ x,y,z >0$, we have to prove $ \sum \frac{z^2}{y^2+5xz} \ge \frac {1}{2}$ By Cauchy-Schwarz inequality: $ \sum \frac{z^2}{y^2+5xz}= \sum \frac{z^4}{y^2z^2+5xz^3} \ge \frac{(x^2+y^2+z^2)^2}{\sum x^2y^2 +5 \sum x^3y}$ It suffits to prove ${ 2(x^2+y^2+z^2)^2 \ge \sum x^2y^2 +5 \sum x^3y}$, which is true by your inequality $ (x^2+y^2+z^2)^2 \ge 3(x^3y+y^3z+z^3x)$ and the inequality $ (x^2+y^2+z^2)^2 \ge 3(x^2y^2+y^2z^2+z^2x^2)$.

Actually if anyone saw my link above it shows that the following one holds for all $ a,b,c,m,n$ satisfying $ abc = 1$ \[ \sum_{cyc}\frac {1}{b(ma + nb)}\ge\frac {3}{m + n} \] And it's easily proven using only Cauchy-Schwarz inequality.

EAsy TRY REARRANGEMENT ineq

PhilAndrew wrote: Let $ a, b, c$ be positive integers for which $ abc = 1$. Prove that $ \sum \frac{1}{b(a+b)} \ge \frac{3}{2}$. Let's write the $LHS$ as $\frac{c^2a}{c^2ab(a+b)}+\frac{b^2c}{b^2ca(c+a)}+\frac{a^2b}{a^2bc(b+c)}$, equivalently $\frac{c^2a}{c(a+b)}+\frac{b^2c}{b(c+a)}+\frac{a^2b}{a(b+c)}$ (since $abc=1$). Then using Cauchy-Schwarz inequality we have $\frac{c^2a}{c(a+b)}+\frac{b^2c}{b(c+a)}+\frac{a^2b}{a(b+c)}\geq \frac{(c\sqrt{a}+b\sqrt{c}+a\sqrt{b})^2}{2(ab+bc+ca)}.$ So, we have to show that $(c\sqrt{a}+b\sqrt{c}+a\sqrt{b})^2\geq 3(ab+bc+ca).$ Expanding and using AM-GM inequality, $abc=1$ we get $(c\sqrt{a}+b\sqrt{c}+a\sqrt{b})^2=c^2a+b^2c+a^2b+2(bc\sqrt{ca}+ca\sqrt{ab}+ab\sqrt{bc})=c^2a+b^2c+a^2b+2(c\sqrt{b}+a\sqrt{c}+b\sqrt{a})=(c^2a+a\sqrt{c}+a\sqrt{c})+(b^2c+c\sqrt{b}+c\sqrt{b})+(a^2b+b\sqrt{a}+b\sqrt{a})\geq 3ca+3bc+3ab=3(ab+bc+ca).$ So, the inequality is proven.

PhilAndrew wrote: Let $ a, b, c$ be positive integers for which $ abc = 1$. Prove that $ \sum \frac{1}{b(a+b)} \ge \frac{3}{2}$. We have \[LHS-RHS=\frac{1}{6(a+b)(b+c)(c+a)}\sum{\frac{(6ab^2c+abc+2bc+3a)(a-1)^2}{ca}}+\frac{1}{2(a+b)(b+c)(c+a)}\sum{\frac{(a-1)^2(b-1)^2}{a}}\ge{0}\]

PhilAndrew wrote: Let $ a, b, c$ be positive integers for which $ abc = 1$. Prove that $ \sum \frac{1}{b(a+b)} \ge \frac{3}{2}$. It was here: http://artofproblemsolving.com/community/c6h81588

nayel wrote: I think you mean positive reals. This is very easy, I can't believe it was proposed in a recent olympiad.. Letting $ a = \frac xy,b = \frac yz, c = \frac zx$ implies \[ LHS = \sum_{cyc}\frac {x^2}{z^2 + xy}\ge\frac {(x^2 + y^2 + z^2)^2}{x^2y^2 + y^2z^2 + z^2x^2 + x^3y + y^3z + z^3x} \]Now it remains to prove that \[ 2(x^2 + y^2 + z^2)^2\ge 3\sum_{cyc}x^2y^2 + 3\sum_{cyc}x^3y \]Which follows by adding the two inequalities \[ x^4 + y^4 + z^4\ge x^3y + y^3z + z^3x \\ \\ \sum_{cyc}(x^4 + x^2y^2)\ge \sum_{cyc}2x^3y \]I bet all the contestants solved this one. EDIT: Here is a related topic: http://www.mathlinks.ro/viewtopic.php?search_id=1111061624&t=165948 i think the last inequality x^4 + y^4 + z^4 >= x^3y +y^3z +z^3x.... (it's true according to rearrangement) but in this question a.b.c = 1 so i think we can't work

ehku wrote: Vasc wrote: This is harder. Let $ a, b, c$ be positive numbers for which $ abc = 1$. Prove that $ \sum \frac {1}{b(5a + b)} \ge \frac {1}{2}$. Put $ a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$ where $ x,y,z >0$, we have to prove $ \sum \frac{z^2}{y^2+5xz} \ge \frac {1}{2}$ By Cauchy-Schwarz inequality: $ \sum \frac{z^2}{y^2+5xz}= \sum \frac{z^4}{y^2z^2+5xz^3} \ge \frac{(x^2+y^2+z^2)^2}{\sum x^2y^2 +5 \sum x^3y}$ It suffits to prove ${ 2(x^2+y^2+z^2)^2 \ge \sum x^2y^2 +5 \sum x^3y}$, which is true by your inequality $ (x^2+y^2+z^2)^2 \ge 3(x^3y+y^3z+z^3x)$ and the inequality $ (x^2+y^2+z^2)^2 \ge 3(x^2y^2+y^2z^2+z^2x^2)$. In your last step Vasc inequality and AM-GM inequality. $\sum_{cyc}x^4+\sum_{cyc}x^2y^2\geq 2\sum_{cyc}x^3y$

nayel wrote: Actually if anyone saw my link above it shows that the following one holds for all $ a,b,c,m,n$ satisfying $ abc = 1$ \[ \sum_{cyc}\frac {1}{b(ma + nb)}\ge\frac {3}{m + n} \]And it's easily proven using only Cauchy-Schwarz inequality. Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, then by Titu $LHS=\sum_{cyc}\frac{z^2}{ny^2+mxz}\geq \frac{(\sum_{cyc}x^2)^2}{n\sum_{cyc}x^2y^2+m\sum_{cyc}x^3y}$ So we have to show that $(m+n)(\sum_{cyc}x^2)^2\geq 3n\sum_{cyc}x^2y^2+3m\sum_{cyc}x^3y$ By Rearrangement inequality, $n(\sum_{cyc}x^2)^2\geq 3n\sum_{cyc}x^2y^2$ By Vasc inequality, $m(\sum_{cyc}x^2)^2\geq 3m\sum_{cyc}x^3y$ So we're done.

nayel wrote: I think you mean positive reals. This is very easy, I can't believe it was proposed in a recent olympiad.. Letting $ a = \frac xy,b = \frac yz, c = \frac zx$ implies \[ LHS = \sum_{cyc}\frac {x^2}{z^2 + xy}\ge\frac {(x^2 + y^2 + z^2)^2}{x^2y^2 + y^2z^2 + z^2x^2 + x^3y + y^3z + z^3x} \]Now it remains to prove that \[ 2(x^2 + y^2 + z^2)^2\ge 3\sum_{cyc}x^2y^2 + 3\sum_{cyc}x^3y \]Which follows by adding the two inequalities \[ x^4 + y^4 + z^4\ge x^3y + y^3z + z^3x \\ \\ \sum_{cyc}(x^4 + x^2y^2)\ge \sum_{cyc}2x^3y \]I bet all the contestants solved this one. EDIT: Here is a related topic: http://www.mathlinks.ro/viewtopic.php?search_id=1111061624&t=165948 How did you come to know that we need to substitute a=x/y at the first place ??

Another one is very nice: a,b,c>0: abc=1 then $$\frac{a}{b(a+c)}+\frac{b}{c(b+a)}+\frac{c}{a(c+b)}\ge\frac{3(ab+bc+ca)}{2(a+b+c)}$$

sol incorrect

Albert123 wrote: The inequality is equivalent: $\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \geq \frac{3}{2}$ we have by AM-AG in LHS: $\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \geq 3\sqrt[3]{\frac{1}{abc(a+b)(b+c)(c+a)}}$ Now, $(a+b)(b+c)(c+a) \geq 8$ $\implies \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \geq 3\sqrt[3]{\frac{1}{(8)}}$ $\implies \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \geq \frac{3}{2}$ $\implies \sum_{cyc}{\frac{1}{b(a+b)}} \ge \frac{3}{2}$.$\blacksquare$

is wrong.

Kgxtixigct wrote: nayel wrote: I think you mean positive reals. This is very easy, I can't believe it was proposed in a recent olympiad.. Letting $ a = \frac xy,b = \frac yz, c = \frac zx$ implies \[ LHS = \sum_{cyc}\frac {x^2}{z^2 + xy}\ge\frac {(x^2 + y^2 + z^2)^2}{x^2y^2 + y^2z^2 + z^2x^2 + x^3y + y^3z + z^3x} \]Now it remains to prove that \[ 2(x^2 + y^2 + z^2)^2\ge 3\sum_{cyc}x^2y^2 + 3\sum_{cyc}x^3y \]Which follows by adding the two inequalities \[ x^4 + y^4 + z^4\ge x^3y + y^3z + z^3x \\ \\ \sum_{cyc}(x^4 + x^2y^2)\ge \sum_{cyc}2x^3y \]I bet all the contestants solved this one. EDIT: Here is a related topic: http://www.mathlinks.ro/viewtopic.php?search_id=1111061624&t=165948 How did you come to know that we need to substitute a=x/y at the first place ?? It's a standard technique to do if you have $abc=1$

Since $abc=1$ we can use the Ravi substitution $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$ where $x,y,z>0$. Now the LHS rewrites as $\sum\limits_{cyc} \frac{y^2 z}{x(xz+y^2)}$, furthermore: \[\sum\limits_{cyc} \frac{y^4z^4}{xy^2z^3(xz+y^2)}\geq \frac{(x^2y^2+y^2z^2+z^2x^2)^2}{\sum\limits_{cyc} x^2y^2z^4 +\sum\limits_{cyc} xy^4z^3}=\frac{\sum\limits_{cyc}x^4y^4+2\sum\limits_{cyc} x^2y^2z^4}{\sum\limits_{cyc}x^2y^2z^4 +\sum\limits_{cyc} xy^4z^3}=F(x,y,z)\]\[F(x,y,z)\geq \frac{3}{2} \iff 2\sum\limits_{cyc}x^4y^4+4\sum\limits_{cyc} x^2y^2z^4\geq 3\sum\limits_{cyc}x^2y^2z^4 +3\sum\limits_{cyc} xy^4z^3\]\[\iff 2\sum\limits_{cyc} x^4y^4+\sum\limits_{cyc} x^2y^2z^4\geq 3\sum\limits_{cyc} xy^4z^3\quad (\star)\]Note that AM-GM implies that $\sum\limits_{cyc} y^4z^4+\sum\limits_{cyc} x^2y^4z^2\geq 2\sum\limits_{cyc} xy^4z^3\quad (1)$ and also that\[4\sum\limits_{cyc} x^4y^4=\sum\limits_{cyc} (x^4y^4+y^4z^4+y^4z^4+y^4z^4)\geq \sum\limits_{cyc} 4xy^4z^3=4\sum\limits_{cyc} xy^4z^3\Longrightarrow \sum\limits_{cyc} x^4y^4\geq \sum\limits_{cyc} xy^4z^3\quad (2)\]Summing $(1)$ and $(2)$ we get the inequality in $(\star)$, so we're done.

Marinchoo wrote: Since $abc=1$ we can use the Ravi substitution $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$ where $x,y,z>0$. Now the LHS rewrites as $\sum\limits_{cyc} \frac{y^2 z}{x(xz+y^2)}$, furthermore: \[\sum\limits_{cyc} \frac{y^4z^4}{xy^2z^3(xz+y^2)}\geq \frac{(x^2y^2+y^2z^2+z^2x^2)^2}{\sum\limits_{cyc} x^2y^2z^4 +\sum\limits_{cyc} xy^4z^3}=\frac{\sum\limits_{cyc}x^4y^4+2\sum\limits_{cyc} x^2y^2z^4}{\sum\limits_{cyc}x^2y^2z^4 +\sum\limits_{cyc} xy^4z^3}=F(x,y,z)\]\[F(x,y,z)\geq \frac{3}{2} \iff 2\sum\limits_{cyc}x^4y^4+4\sum\limits_{cyc} x^2y^2z^4\geq 3\sum\limits_{cyc}x^2y^2z^4 +3\sum\limits_{cyc} xy^4z^3\]\[\iff 2\sum\limits_{cyc} x^4y^4+\sum\limits_{cyc} x^2y^2z^4\geq 3\sum\limits_{cyc} xy^4z^3\quad (\star)\]Note that AM-GM implies that $\sum\limits_{cyc} y^4z^4+\sum\limits_{cyc} x^2y^4z^2\geq 2\sum\limits_{cyc} xy^4z^3\quad (1)$ and also that\[4\sum\limits_{cyc} x^4y^4=\sum\limits_{cyc} (x^4y^4+y^4z^4+y^4z^4+y^4z^4)\geq \sum\limits_{cyc} 4xy^4z^3=4\sum\limits_{cyc} xy^4z^3\Longrightarrow \sum\limits_{cyc} x^4y^4\geq \sum\limits_{cyc} xy^4z^3\quad (2)\]Summing $(1)$ and $(2)$ we get the inequality in $(\star)$, so we're done. wow! solution!