\[ \sum_{cyc} \frac{x^5}{x^5+y^2+z^2} \geq \sum_{cyc} \frac{x^2}{x^5+y^2+z^2} \] \[ \sum_{cyc} \left( 1 - \frac{y^2+z^2}{x^5+y^2+z^2} \right) \geq \sum_{cyc} \frac{x^2}{x^5+y^2+z^2} \] \[ 3 \geq (x^2+y^2+z^2)\cdot \left( \sum_{cyc} \frac{1}{x^5+y^2+z^2} \right) \] It is clear we only need to investigate the case xyz = 1. After substitution x=a^2/bc, y=b^2/ac, z=c^2/ca, we get an expression, in which we now simplify putting a^3-->a, b^3-->b, c^3-->c, which is bijective everywhere. This results in: \[ 3 \geq (a^3bc+ab^3c+abc^3)( \frac{1}{a^5+ab^3c+abc^3} + \frac{1}{a^3bc+b^5+abc^3} + \frac{1}{a^3bc+ab^3c+c^5} )\] Working out gives: $\sum_{cyc} a^{10}bc + 2 \sum_{sym} a^7b^5 + \sum_{cyc}a^6b^3c^3 \geq \sum_{sym} a^6b^5c + \sum_{cyc}a^8b^2c^2 +\sum_{cyc}a^5b^5c^2 + \sum_{sym} a^6b^4c^2 $ Which follows trivially from $\sum_{cyc}a^{10}bc+\sum_{cyc}a^6b^3c^3\geq2\sum_{cyc}a^8b^2c^2$ $\sum_{cyc}a^8b^2c^2\geq\frac12\sum_{sym}a^6b^4c^2$ $\sum_{sym}a^7b^5\geq\sum_{sym}a^6b^5c$ $\frac12\sum_{sym}a^7b^5\geq\sum_{cyc}a^5b^5c^2$ and $\frac12\sum_{sym}a^7b^5\geq\frac12\sum_{sym}a^6b^4c^2$ This proves everyone can solve a number 3 these days. Image not found

A solution using Cauchy's inequality: \[ \frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}-y^{2}}{x^{2}+y^{5}+z^{2}}+\frac{z^{5}-z^{2}}{x^{2}+y^{2}+z^{5}}\geq 0\] is equivalent to : \[ \frac{1}{x^{5}+y^{2}+z^{2}}+\frac{1}{x^{2}+y^{5}+z^{2}}+\frac{1}{x^{2}+y^{2}+z^{5}}\leq \frac{3}{x^{2}+y^{2}+z^{2}}\] Now, from Cauchy's inequality we have: \[( x^{2}\cdot x^{3}+y^{2}+z^{2}) ( x^{2}\cdot \frac{1}{x^{3}}+y^{2}+z^{2}) \geq (x^{2}+y^{2}+z^{2})^{2}\] But from the given condition we have $ x^{2}\cdot \frac{1}{x^{3}}= \frac{1}{x}\leq yz$ so we obtain that: \[(x^{5}+y^{2}+z^{2}) (yz+y^{2}+z^{2}) \geq ( x^{2}\cdot x^{3}+y^{2}+z^{2}) ( x^{2}\cdot \frac{1}{x^{3}}+y^{2}+z^{2}) \geq (x^{2}+y^{2}+z^{2})^{2}\] All that is left now is to notice that \[ \sum_{\textrm{cyc}}\frac{1}{x^{5}+y^{2}+z^{2}}\leq \sum_{\textrm{cyc}}\frac{yz+y^{2}+z^{2}}{(x^{2}+y^{2}+z^{2})^{2}}\leq \sum_{\textrm{cyc}}\frac{\frac{y^{2}+z^{2}}{2}+y^{2}+z^{2}}{(x^{2}+y^{2}+z^{2})^{2}}= \frac{3}{x^{2}+y^{2}+z^{2}}\]

Same solution as mine, andrei , you beat me to the bunch. Cauchying the $x^5 + y^2 + z^2$ bottom is the key, and a nice trick. I was able to see this trick fairly easily, but I have to admit that I had some recent inspiration. Check out this nice problem to see where the inspiration was from.

Edit: as blahblahblah points out the convexity argument is in the wrong direction, so it does not quite work as written. I do not see at the moment whether the lower bound stated below is correct.

By the way, this calculation gives a lower bound of $\frac{3t}{1+t} \geq 0$ where $t = (xyz-1)/(x^2+y^2+z^2)$.

Unless I'm missing something, shouldn't your sign be the other way for convexity? We get $\sum_i \frac {1}{1+w_i}\geq \frac {3}{1+\sum_i w_i/3}\leq 3$, which isn't helpful...

Peter VDD, you beat me in posting the ultimate solution! MUIRHEAD ROCKS!!!

georgesand wrote: Peter VDD, you beat me in posting the ultimate solution! MUIRHEAD ROCKS!!! Well it isn't purely muirhead... just a brute force muirhead on the start will most likely bring you nowhere, you need to do the manipulation there first [they're not all muirhead btw... like the first one of the last post is AM-GM ] [Moderator edit. The "Muirhead inequality" referred to here is also called the "bunching inequality". Further discussion of the use of this inequality was moved into http://www.mathlinks.ro/Forum/viewtopic.php?t=102840 .]

I'm wrong

A function, which is positive and tends to 0 as $x$ goes to infinity may not be concave.

Peter VDD, you we're faster in posting than me, too.. my solution is essentially the same, but it has different 'combinatorics' using xyz=1 I supstitute every a^2 with the a^3 bc and then multyply everything to get your inequality: 2 * [7,5,0] + [10, 1, 1] + [6, 3, 3] >= [5, 5, 2] + [6, 5, 1] + [6, 4, 2] + [8, 2, 2]. My proof is a bit different than yours' : [10, 1, 1] + [6, 3, 3) >= 2 [8, 2, 2] by AG mean inequality, and the rest are by murhead: [8, 2, 2]>= [5, 5, 2] => [10, 1, 1] + [6, 3, 3] >= [8, 2, 2] + [5, 5, 2] [7, 5, 0] >= [6, 5, 1] [7, 5, 0] >= [6, 4, 2] collecting the last three gives the desired result. Really happy about beating the 3rd problem!

This problem is far too easy for a number 3. At a first glance, it looks a bit scary because of the fifth powers. But if you start working on it, straightforward methods can be used to solve it. Actually, problem 2 is also far too easy for a problem 2. .

Putting $\displaystyle\sum\frac{x^5-x^2}{x^5+y^2+z^2}\ge \displaystyle\sum\frac{2x^2-y^2-z^2}{2x^2+2y^2+2z^2}=0\Leftrightarrow \vskip 0,2in 3x^5y^2+3x^5z^2+y^4+z^4+2y^2z^2\ge 4x^2y^2+4x^2z^2+2x^4$ the last inequality can be obtained by using AM-GM and $xyz\ge1,$ $ x^5y^2+x^5z^2\ge 2x^4,x^5y^2+x^5y^2+y^4+y^2z^2\ge 4x^2y^2,\vskip 0,1in x^5z^2+x^5z^2+z^4+y^2z^2\ge 4x^2z^2 $

May I ask: Who is Hojoo Lee? He proposed two problems last year and one this year...... A great poster.

Shalom everybody. i am new to this forum but i heard a lot of good things about this place from my friend. Well, let me give a simple solution. It is so simple i have troubles beleiving myself. Maybe someone will find a mistake firstly, x^2+y^2+z^2 >= xy+xz+yz = xyz (1/x+1/y+1/z) >= 1/x+1/y+1/z Then we obtain an auxiliary inequality. Namely, (x^5-x^2)/(x^5+y^2+z^2) >= (x^5-x^2)/(x^5+ x^3*y^2+x^3*z^2) (which is, of cours = (x^2-1/x)/(x^2+y^2+z^2) ) The last inequality i shall explain. If x<1 then multypliyng below by x^3 reduces the denominator, so increases the abs. valueof the number so decreases the number, since the number is negative. The second case: x>1. Then the denominator is increased, so the number is reduced, cause it is positive. Summing up inequalities of theat kind, we get LHS >=(x^2-1/x+y^2-1/y+z^2-1/z)/(x^2+y^2+z^2) >=0 because of the first inequality i wrote. That's it.

I had a similar idea : Obviuosly $(x^2+y^2+z^2)x^3-(x^5+y^2+z^2)$ and $x^2- \dfrac 1x$ have the same sign; hence \[ \frac {x^5-x^2}{x^5+y^2+z^2} \geq \frac {x^2- \dfrac 1x}{x^2+y^2+z^2} \geq \frac {x^2-yz}{x^2+y^2+z^2} \] add the three similar inequalities to get the result.

My proof is quite simple, using the fact that $(x^5+y^2+z^2)(\frac{1}{x} + y^2+z^2) \geq (x^2+y^2+z^2)^2$

treegoner wrote: My proof is quite simple, using the fact that $(x^5+y^2+z^2)(\frac{1}{x} + y^2+z^2) \geq (x^2+y^2+z^2)^2$ Well not everybody sees that the first time

Beautiful

Valentin Vornicu wrote: Let $x,y,z$ be three positive reals such that $xyz\geq 1$. Prove that \[ \frac { x^5-x^2 }{x^5+y^2+z^2} + \frac {y^5-y^2}{x^2+y^5+z^2} + \frac {z^5-z^2}{x^2+y^2+z^5} \geq 0 . \]Hojoo Lee, Korea $\begin{aligned} \frac { x^5-x^2 }{x^5+y^2+z^2} + \frac {y^5-y^2}{x^2+y^5+z^2} + \frac {z^5-z^2}{x^2+y^2+z^5} &=\dfrac{x^3\left( x-\frac 1x\right)}{x^3\cdot x^2+y+z}+\dfrac{y^3\left( y-\frac 1y\right)}{y^3\cdot y^2+z+x}+\dfrac{z^3\left( z-\frac 1z\right)}{z^3\cdot z^2+x+y}\\ &=\dfrac{x^2-\frac 1x}{x^2+\frac{y^2+z^2}{x^3}}+\dfrac{y^2-\frac 1y}{y^2+\frac{z^2+x^2}{y^3}}+\dfrac{z^2-\frac 1z}{z^2+\frac{x^2+y^2}{z^3}}\\ &\geqslant\dfrac{x^2-\frac 1x}{x^2+y^2+z^2}+\dfrac{y^2-\frac 1y}{y^2+z^2+z^2}+\dfrac{z^2-\frac 1z}{z^2+x^2+y^2}\\ &=\dfrac{x^2-\frac 1x+y^2-\frac 1y+z^2-\frac 1z}{x^2+y^2+z^2}\\ &\geqslant\frac{x^2+y^2+z^2-xy-yz-zx}{x^2+y^2+z^2}\\ &\geqslant 0. \end{aligned}$

Valentin Vornicu wrote: Let $x,y,z$ be three positive reals such that $xyz\geq 1$. Prove that \[ \frac { x^5-x^2 }{x^5+y^2+z^2} + \frac {y^5-y^2}{x^2+y^5+z^2} + \frac {z^5-z^2}{x^2+y^2+z^5} \geq 0 . \]Hojoo Lee, Korea Step 1: We prove that $\frac { x^5}{x^5+y^2+z^2} + \frac {y^5}{x^2+y^5+z^2} + \frac {z^5}{x^2+y^2+z^5}\geqslant 1.$ $\frac { x^5}{x^5+y^2+z^2}\geqslant\frac{x^5}{x^5+(y^2+z^2)xyz}=\frac{x^4}{x^4+y^3z+z^3y}\geqslant\frac{x^4}{x^4+y^4+z^4}.$ $\frac { y^5}{y^5+z^2+x^2}\geqslant\frac{y^5}{y^5+(z^2+x^2)yzx}=\frac{y^4}{y^4+z^3x+x^3z}\geqslant\frac{y^4}{y^4+z^4+x^4}.$ $\frac { z^5}{z^5+x^2+y^2}\geqslant\frac{z^5}{z^5+(x^2+y^2)zxy}=\frac{z^4}{z^4+x^3y+y^3x}\geqslant\frac{z^4}{z^4+x^4+y^4}.$ $\therefore\frac { x^5}{x^5+y^2+z^2} + \frac {y^5}{x^2+y^5+z^2} + \frac {z^5}{x^2+y^2+z^5}\geqslant \frac{x^4}{x^4+y^4+z^4}+\frac{y^4}{y^4+z^4+x^4}+\frac{z^4}{z^4+x^4+y^4}=1.$ Step 2: We prove that $\frac { x^2}{x^5+y^2+z^2} + \frac {y^2}{x^2+y^5+z^2} + \frac {z^2}{x^2+y^2+z^5}\leqslant 1.$ $\frac { x^2}{x^5+y^2+z^2}\leqslant\frac { x^2\cdot xyz}{x^5+xyz(y^2+z^2)}=\frac{x^2yz}{x^4+y^3z+z^3y}\leqslant\frac{x^2yz}{x^2yz+y^2zx+z^2xy}=\frac{x}{x+y+z}.$ $\frac { y^2}{y^5+z^2+x^2}\leqslant\frac { y^2\cdot yzx}{y^5+yzx(z^2+x^2)}=\frac{y^2zx}{y^4+z^3x+x^3z}\leqslant\frac{y^2zx}{y^2zx+z^2xy+x^2yz}=\frac{y}{y+z+x}.$ $\frac { z^2}{z^5+x^2+y^2}\leqslant\frac { z^2\cdot zxy}{z^5+zxy(x^2+y^2)}=\frac{z^2xy}{z^4+x^3y+y^3x}\leqslant\frac{z^2xy}{z^2xy+x^2yz+y^2zx}=\frac{z}{z+x+y}.$ $\therefore\frac { x^2}{x^5+y^2+z^2} + \frac {y^2}{x^2+y^5+z^2} + \frac {z^2}{x^2+y^2+z^5}\leqslant\frac{x}{x+y+z}+\frac{y}{y+z+x}+\frac{z}{z+x+y}=1.\blacksquare$

We have $$ \sum \frac{x^5}{x^5+y^2+z^2}\\ \ge \frac{(\sum x^3)^2}{\sum x^6 + xy^2 + xz^2} = \frac{\sum x^6 + 2\sum x^3y^3}{\sum x^6 + \sum xy^2}\\ \ge \frac{\sum x^6 + 2\sum x^3y^3}{\sum x^6+\sum x^2y^3z}\\ \ge 1$$$$\ge \frac{2\sum x^2yz+\sum x^2y^2}{(x^2+y^2+z^2)^2} \ge \sum \frac{x^2(\frac{1}{x} +y^2+z^2)}{(x^2+y^2+z^2)^2} \ge \sum \frac{x^2}{x^5+y^2+z^2}$$ Where the first step is Titu's lemma, the last step is Cauchy, and the fact that $xyz\ge 1$ is used in some steps. Some steps also use Muirhead's

We want to prove $$\sum_{cyc}\frac{x^2+y^2+z^2}{x^5+y^2+z^2}\le 3.$$(We're motivated to do this because the fifth power is really out of place.) Indeed, note that $$(x^5+y^2+z^2)(yz+y^2+z^2)\ge (x^2+y^2+z^2)^2\iff\frac{x^2+y^2+z^2}{x^5+y^2+z^2}\le\frac{yz+y^2+z^2}{x^2+y^2+z^2}\implies LHS\le \sum_{cyc}\frac{yz+y^2+z^2}{x^2+y^2+z^2}\le 3\iff xy+yz+zx\le x^2+y^2+z^2,$$which is obviously true. $\blacksquare$

\[ \sum_{cyc} \frac{x^5}{x^5+y^2+z^2} \geq \sum_{cyc} \frac{x^2}{x^5+y^2+z^2} \]\[ \sum_{cyc} \left( 1 - \frac{y^2+z^2}{x^5+y^2+z^2} \right) \geq \sum_{cyc} \frac{x^2}{x^5+y^2+z^2} \]\[ 3 \geq (x^2+y^2+z^2)\cdot \left( \sum_{cyc} \frac{1}{x^5+y^2+z^2} \right) \]\[ \frac{1}{x^{5}+y^{2}+z^{2}}+\frac{1}{x^{2}+y^{5}+z^{2}}+\frac{1}{x^{2}+y^{2}+z^{5}}\leq \frac{3}{x^{2}+y^{2}+z^{2}}\]$\newline$ From Cauchy-Schwarz and $1/x \le yz$ we get: \[(x^{5}+y^{2}+z^{2}) (yz+y^{2}+z^{2}) \geq ( x^{2}\cdot x^{3}+y^{2}+z^{2}) ( x^{2}\cdot \frac{1}{x^{3}}+y^{2}+z^{2}) \geq (x^{2}+y^{2}+z^{2})^{2}\] \[ \sum_{cyc}\frac{1}{x^{5}+y^{2}+z^{2}}\leq \sum_{cyc}\frac{yz+y^{2}+z^{2}}{(x^{2}+y^{2}+z^{2})^{2}}\leq \sum_{cyc}\frac{\frac{y^{2}+z^{2}}{2}+y^{2}+z^{2}}{(x^{2}+y^{2}+z^{2})^{2}}= \frac{3}{x^{2}+y^{2}+z^{2}}\]

Comment1 on gen1

Comment2 on gen1

Gen0 comment

Gen1 itself

Note that Cauchy gives us \[(x^5+y^2+z^2)(yz+y^2+z^2) \ge (x^2 \sqrt{xyz}+y^2+z^2)^2 \ge (x^2+y^2+z^2)^2,\] so we have \[\sum_{\text{cyc}} \frac{1}{x^5+y^2+z^2} \leq \sum_{\text{cyc}} \frac{yz+y^2+z^2}{(x^2+y^2+z^2)^2} \leq \frac{3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^2} = \sum_{\text{cyc}} \frac{1}{x^2+y^2+z^2}.\] Shifting everything to the RHS and multiplying by $x^2+y^2+z^2$, we get the desired. $\blacksquare$

This can be rearranged into $\frac1{x^5+y^2+z^2}+\frac1{y^5+x^2+z^2}+\frac1{z^5+x^2+y^2}\le\frac3{x^2+y^2+z^2},$ which becomes $\frac3{x^2+y^2+z^2}-\frac{xy}{z^4+x^3y+y^3x}-\frac{yz}{x^4+y^3z+z^3y}-\frac{xz}{y^4+x^3z+z^3x}\ge0$ after homogenizing. Now cross multiplying everything and simplifying gives \[\frac12\sum_{\text{sym}}x^8y^2z^2+2\sum_{\text{sym}}x^7y^5-\sum_{\text{sym}}x^6y^5z-\sum_{\text{sym}}x^6y^4z^2-\frac12\sum_{\text{sym}}x^5y^5z^2+\sum_{\text{cyc}}x^6yz(x^2-yz)^2\ge0,\]which follows from Muirhead.

Gen1 proof

Note that $$x^2 \leq x^3yz\leq \frac{x^3}2(y^2+z^2) \quad \text{ and } \quad y^2+z^2\leq xyz(y^2+z^2)\leq \frac x2 (y^2+z^2)^2.$$So, \(LHS\geq \sum_{cyc} \frac{x^2(2x^2-y^2-z^2)}{2x^4+(y^2+z^2)^2}\). We let $(x,y,z)=(\sqrt a,\sqrt b,\sqrt c)$ and assume wlog that $a+b+c=1$. Let \(f(x)=\frac{x(3x-1)}{2x^2+(1-x)^2}\) then we want to prove $f(a)+f(b)+f(c)\geq 0$. Now by the tangent line trick we only have to prove $$f(x)\geq f(1/3)+f'(1/3)(x-1/3)\iff \frac{x(3x-1)}{2x^2+(1-x)^2}\geq \frac 32\left (x-\frac 13\right ) \iff (1-x)(3x-1)^2\geq 0$$which is obviously true.

Please contact westskigamer@gmail.com if there is an error with my solution.