frenchy wrote: Given real numbers $x,y,z$ such that $x+y+z=0$, show that \[\dfrac{x(x+2)}{2x^2+1}+\dfrac{y(y+2)}{2y^2+1}+\dfrac{z(z+2)}{2z^2+1}\ge 0\] When does equality hold? WLOG: $x^2\le y^2\le z^2$ $\left(\frac{x(x+2)}{2x^2+1}+\frac{1}{2}\right)+\left(\frac{y(y+2)}{2y^2+1}+\frac{1}{2}\right)$ $=\frac{(2x+1)^2}{2\left(2x^2+1\right)}+\frac{(2y+1)^2}{2\left(2y^2+1\right)}$ $\ge\frac{\left((2x+1)+(2y+1)\right)^2}{2\left(2x^2+1\right)+2\left(2y^2+1\right)}$ $=\frac{(1-z)^2}{x^2+y^2+1}$ $\ge\frac{(1-z)^2}{2z^2+1}$ $=1-\frac{z(z+2)}{2z^2+1}$

mahanmath wrote: Same as mine abch42 , I like to add equality cases here: $(x,y,z) = (\frac{-1}{2} , \frac{-1}{2} , 1)$ up to permutation . and $(x,y,z)=(0,0,0)$

Nice Problem and amazing solution. I would like to add my solution because i think its correct. Firstly, we are clearing the denominators. Then, it is enough to prove that 2( Σcyc x^2 y^2 z ) + Σcyc x^2 y >= 0 . Then, we separate these 12 terms to 6 and 6 terms ( symmetrically ). Finally, we apply the AM-GM inequality to both groups and we are done. The equality occurs when x,y,z are 0,0,0, and 1,-1/2,-1/2 with all permutations.

frenchy wrote: Given real numbers $x,y,z$ such that $x+y+z=0$, show that \[\dfrac{x(x+2)}{2x^2+1}+\dfrac{y(y+2)}{2y^2+1}+\dfrac{z(z+2)}{2z^2+1}\ge 0\] When does equality hold? Notice that $\frac{x(x+2)}{2x^2+1}=\frac{(2x+1)^2}{2(2x^2+1)}-\frac{1}{2}.$ So the inequality can be written as $\sum \frac{(2x+1)^2}{2x^2+1} \ge 3.$ Now, using the Cauchy-Schwarz inequality, we see that $2x^2=\frac{4}{3}x^2+\frac{2}{3}(y+z)^2 \le \frac{4}{3}x^2+\frac{4}{3}(y^2+z^2).$ Therefore, $\sum \frac{(2x+1)^2}{2x^2+1} \ge 3\sum \frac{(2x+1)^2}{4(x^2+y^2+z^2)+3}=3.$

kalifornia001 wrote: I have done in a different way by replacing $0$ with $x+y+z$, then find the same denominator. And in the end i got the desired result. $\sum{\frac{x(x+2)}{2x^2+1}}\geq\frac{2}{3}(x+y+z)\sum{\frac{1-5y-5z}{1+2x^2}}$ holds for all real numbers $x,y,z.$ CDP100 wrote: Would you show us your solution?

Another solution: Multiplying by $(2a^2+1)(2b^2+1)(2c^2+1)$ and expanding we have to prove: $8abc(ab+bc+ca) + 4\sum_{cyclic}{a(b^2 + c^2)} + 12a^2b^2c^2 + a^2 + b^2 + c^2 + 4(a^2b^2 + b^2c^2 + c^2a^2) = 8abc(ab+bc+ca) + 4\sum_{cyclic}{bc(b + c)} + 12a^2b^2c^2 + a^2 + b^2 + c^2 + 4(a^2b^2 + b^2c^2 + c^2a^2) = 8abc(ab+bc+ca) -12abc + 12a^2b^2c^2 + a^2 + b^2 + c^2 + 4(a^2b^2 + b^2c^2 + c^2a^2) - 4abc(a + b + c) = \sum_{cyclic}{(2ab + 2abc - c)^2} \geq 0$ which is obviously true. To have equality, we must have $2ab + 2abc - c = 0 \Rightarrow 2abc + 2abc \times c - c^2 = 0$ and simillarly for the other two variables, hence the three variables must be roots of the 2-degree monic polyomial: $x^2 - 2kx - 2k = 0, k=abc$, and hence two of them must be equal: Assume WLOG: $(a,b,c) = (m,m,-2m)$, then $2ab + 2abc = c \Rightarrow 2m^2 - 4m^3 = -2m \Rightarrow m \in \{0,-\frac{1}{2}\,1\}$, and by checking we can see that equality holds only for $m \in \{0,-\frac{1}{2}\}$ Hence we have equality for $(a,b,c) = (0,0,0)$ or $(\frac{-1}{2}, \frac{-1}{2},1)$ and it's permutations.

I have an ugly solution, but I will post it anyway. Out of the three numbers it is always possible to pick two of them, such that, the two are both non-negative or both non-positive. WLOG let $x,y$ be those two numbers. (It follows $xy \geq 0$) First one substitutes $z=-(x+y)$ . We get $ \frac{x^2+2x}{2x^{2}+1}+\frac{y^2+2y}{2y^{2}+1}+\frac{(x+y)^2}{2(x+y)^{2}+1}\ge \frac{2(x+y)}{2(x+y)^{2}+1} $ Since $ 2(x+y)^{2}+1>0$ this is equivalent to : $ (x^2+2x) \frac{2(x+y)^{2}+1}{2x^{2}+1} + (y^2+2y)\frac{2(x+y)^{2}+1}{2y^{2}+1}+ (x+y)^2\frac{2(x+y)^{2}+1}{2(x+y)^{2}+1}\ge 2(x+y) $ $ (x^2+2x) \frac{2x^2+1+4xy+2y^2}{2x^{2}+1} + (y^2+2y)\frac{2y^2+1+4xy+2x^2}{2y^{2}+1}+ (x+y)^2\ge 2(x+y) $ $ (x^2+2x) \frac{4xy+2y^2}{2x^{2}+1} + (y^2+2y)\frac{4xy+2x^2}{2y^{2}+1}+ (x+y)^2+ y^2+x^2\ge 0 $ We now prove $\frac{x^2+2x}{2x^{2}+1} \ge -\frac{1}{2}$ It is possible to change it into this form $4x^2+4x+1\ge 0$ and one can then use the standard formulae to calculate that the minimum of that happens at $x=-\frac{1}{2}$ and is equal to zero. Since $x,y$ have the same sign, $4xy+2y^2 \ge 0$ and $4xy+2x^2 \ge 0$ Therefore one can estimate the inequality in this way: $ (x^2+2x) \frac{4xy+2y^2}{2x^{2}+1} + (y^2+2y)\frac{4xy+2x^2}{2y^{2}+1}+ (x+y)^2+ y^2+x^2 \ge $ $\ge -(2xy+y^2)-(2xy+x^2) + (x+y)^2+ y^2+x^2 = (x-y)^2 \ge 0 $ So the inequality is proved. For equality we know that $x=y$ and either $x=-\frac{1}{2}$ or $4xy+2x^2=6x^2=0$. This gives us two equality cases $(-\frac{1}{2},-\frac{1}{2},1)$ with all its permutations and $(0,0,0)$

Generalization Given real numbers $x,y,z$ such that $x+y+z=0$ and $a,b,c,d,p>0,{{a}^{2}}d\ge 2{{b}^{2}}c,p=\frac{2abd}{{{a}^{2}}d-{{b}^{2}}c}$, show that \[\frac{x(x+p)}{c{{x}^{2}}+d}+\frac{y(y+p)}{c{{y}^{2}}+d}+\frac{z(z+p)}{c{{z}^{2}}+d}\ge 0.\] When does equality hold?

AwesomeToad wrote:

^^ Seems correct, but I'm kind of surprised no one has used it yet. Is it correct? No,because $4x^3+2x<0,if x<0$

AwesomeToad wrote: That shouldn't affect the correctness of my application of Cauchy/Titu's Lemma. If some denominator is less than $0$, then it doesn't work in general, for example: \[-5=\frac{(6)^2}{-4}+\frac{(2)^2}{2}+\frac{(4)^2}{8}\ge \frac{(12)^2}{6}=24\] is clearly false. (As for the reasoning behind this, its because the proof of Cauchy form $\sum\frac{a_i^2}{b_1}\ge \frac{(\sum a_i)^2}{\sum b_i}$ relies on $\sqrt{b_i}$ being real for each $i$.)

WLOG: $x^2\le y^2\le z^2$ $x(x+2)/(2x^2+1) \geq x(x+2)/(2z^2+1)$ and .... so LHS $\geq (x^2+y^2+z^2)/(2x^2+1) \geq 0$ i think it's wrong but i dont find the mistake :p

Vasc wrote: This is also nice. If $a,b,c,d$ are real numbers such that $a+b+c+d=0$, then \[\frac{(a-1)^2}{3a^2+1}+\frac{(b-1)^2}{3b^2+1}+\frac{(c-1)^2}{3c^2+1}+\frac{(d-1)^2}{3d^2+1}\le 4.\] Since $ \dfrac{(a-1)^2}{3a^2+1}=\dfrac{4}{3}-\dfrac{(3a+1)^2}{3(3a^2+1)} $ We can rewrite the inequality as the follows $ \sum \dfrac{(3a+1)^2}{3a^2+1} \geq 4 $ Because $ 3a^2=\dfrac{9}{4}a^2+\dfrac{3}{4}(b+c+d)^2 \leq \dfrac{9}{4}(a^2+b^2+c^2+d^2) $ So $ \sum \dfrac{(3a+1)^2}{3a^2+1} \geq 4 \sum \dfrac{(3a+1)^2}{9(a^2+b^2+c^2+d^2) +4}=4 $

frenchy wrote: Given real numbers $x,y,z$ such that $x+y+z=0$, show that \[\dfrac{x(x+2)}{2x^2+1}+\dfrac{y(y+2)}{2y^2+1}+\dfrac{z(z+2)}{2z^2+1}\ge 0\] When does equality hold? $x^2=(y+z)^2\leq2(y^2+z^2)\implies 3x^2\leq2(x^2+y^2+z^2),$ $\dfrac{(2x+1)^2}{2x^2+1}+\dfrac{(2y+1)^2}{2y^2+1}+\dfrac{(2z+1)^2}{2z^2+1}$ $\geq\dfrac{3(2x+1)^2}{4(x^2+y^2+z^2)+3}+\dfrac{3(2y+1)^2}{4(x^2+y^2+z^2)+3}+\dfrac{3(2z+1)^2}{4(x^2+y^2+z^2)+3}=3,$ $\implies \dfrac{(2x+1)^2}{2x^2+1}+\dfrac{(2y+1)^2}{2y^2+1}+\dfrac{(2z+1)^2}{2z^2+1}\ge 3$ $\iff\dfrac{x(x+2)}{2x^2+1}+\dfrac{y(y+2)}{2y^2+1}+\dfrac{z(z+2)}{2z^2+1}\ge 0$

sqing wrote: Given real numbers $x,y,z$ such that $x+y+z=0$, show that$$ \frac{x+1}{x^2+3}+ \frac{y+1}{y^2+3}+ \frac{z+1}{z^2+3} \le 1.$$ the given inequality is equivalent to $$\sum_{cyc}\frac{3(x+1)}{x^2+3}\le 3$$ $$\sum_{cyc}\frac{x^2-3x}{x^2+3}\ge 0$$ $$\sum_{cyc}\frac{(x-1)^2}{x^2+3}\ge 1$$ which is true since $x^2\le x^2+y^2+z^2$ so $$\sum_{cyc}\frac{(x-1)^2}{x^2+3}\ge \sum_{cyc}\frac{(x-1)^2}{x^2+y^2+z^2+3}=1$$

$ \frac{x+1}{x^2+3}+ \frac{y+1}{y^2+3}+ \frac{z+1}{z^2+3} \le 1\iff \frac{(x-1)^2}{x^2+3}+\frac{(y-1)^2}{y^2+3}+\frac{(z-1)^2}{z^2+3}\ge 1$, $x^2\le x^2+y^2+z^2\implies \sum_{cyc}\frac{(x-1)^2}{x^2+3}\ge \sum_{cyc}\frac{(x-1)^2}{x^2+y^2+z^2+3}=1$. Nice. Thanks.

Very beautiful problem .

leonardg wrote: Very beautiful problem . Yeah . Generalization of Balkan 2011

Denote $f(x)=\frac{x(x+2)}{2x^2+1}$ Notice that $f''(x) \geq 0$ $\Rightarrow f$ is convex By Jensen inequality we have $f(x)+f(y)+f(z) \geq 3f(\frac{x+y+z}{3})=3f(0)=0$ so $f(x)+f(y)+f(z) \geq0$ with equality if and only if $x=y=z=0$ If there is anything wrong with this solution please inform me.

Mr.Chem-Mathy wrote: Denote $f(x)=\frac{x(x+2)}{2x^2+1}$ Notice that $f''(x) \geq 0$ $\Rightarrow f$ is convex By Jensen inequality we have $f(x)+f(y)+f(z) \geq 3f(\frac{x+y+z}{3})=3f(0)=0$ so $f(x)+f(y)+f(z) \geq0$ with equality if and only if $x=y=z=0$ If there is anything wrong with this solution please inform me. How did you arrive at the conclusion that $f$ is convex? It seems to me that it definitely is not convex everywhere. (Specifically $x \leq -1$ seems to not work.)

mathwizn wrote: Mr.Chem-Mathy wrote: Denote $f(x)=\frac{x(x+2)}{2x^2+1}$ Notice that $f''(x) \geq 0$ $\Rightarrow f$ is convex By Jensen inequality we have $f(x)+f(y)+f(z) \geq 3f(\frac{x+y+z}{3})=3f(0)=0$ so $f(x)+f(y)+f(z) \geq0$ with equality if and only if $x=y=z=0$ If there is anything wrong with this solution please inform me. How did you arrive at the conclusion that $f$ is convex? It seems to me that it definitely is not convex everywhere. (Specifically $x \leq -1$ seems to not work.) Because the second derivative is a constant which is greater than 0. If $f$ were concave I think we would get a contradiction to the problem. I probably differentiated the function wrong though since I've never taken a class of calc/pre-calc. Thanks for pointing it out though; I'll look deeper into it.

Mr.Chem-Mathy wrote: mathwizn wrote: Mr.Chem-Mathy wrote: Denote $f(x)=\frac{x(x+2)}{2x^2+1}$ Notice that $f''(x) \geq 0$ $\Rightarrow f$ is convex By Jensen inequality we have $f(x)+f(y)+f(z) \geq 3f(\frac{x+y+z}{3})=3f(0)=0$ so $f(x)+f(y)+f(z) \geq0$ with equality if and only if $x=y=z=0$ If there is anything wrong with this solution please inform me. How did you arrive at the conclusion that $f$ is convex? It seems to me that it definitely is not convex everywhere. (Specifically $x \leq -1$ seems to not work.) Because the second derivative is a constant which is greater than 0. If $f$ were concave I think we would get a contradiction to the problem. I probably differentiated the function wrong though since I've never taken a class of calc/pre-calc. Thanks for pointing it out though; I'll look deeper into it. http://www.wolframalpha.com/input/?i=second+derivative+of+%5Cfrac%7Bx(x%2B2)%7D%7B2x%5E2%2B1%7D And no, the fact that the function is concave for some intervals doesn't contradict the problem's statement.

Alright, thanks. @everyone, ignore my solution.

ryuzaki wrote: WLOG: $x^2\le y^2\le z^2$ $x(x+2)/(2x^2+1) \geq x(x+2)/(2z^2+1)$ and .... so LHS $\geq (x^2+y^2+z^2)/(2x^2+1) \geq 0$ i think it's wrong but i dont find the mistake :p Hello, here I am, browsing the forum six years later, as I stumble upon your solution. I doubt you don't already know what went wrong with your attempt, but I'll point it out anyway: x can be a negative number, and so $x(x+2)$ can be negative, and so if you work with inequalities beneath it the sign will be reversed. Sorry for the late response for your probably already answered question, good day!

It looks easy by $AM\ge HM$ Giving it shortly

paragdey01 wrote: It looks easy by $AM\ge HM$ It looks the same with or without AM-HM. Meaning it looks difficult . Please give a shot with AM-HM .

who has solved this propblem with function

Note that $\frac{x^2+2x}{2x^2+1} = \frac{\frac{(2x+1)^2}{2x^2+1}-1}{2}$ so we need to prove $\sum \frac{(2x+1)^2}{2x^2+1} \ge 3$. Note that $x = -(y+z)$ so $x^2 = (y+z)^2 \le 2(y^2 + z^2)$ so $\sum \frac{(2x+1)^2}{2x^2+1} = 3\sum \frac{(2x+1)^2}{4x^2 + 2x^2 +3} \ge 3\sum \frac{(2x+1)^2}{4x^2 + 4y^2 + 4z^2 +3} = 3$.

Excellent

Multiply both sides by 2 getting cyc sum of (2x^2+4x)/(2x^2+1)>= 0 cyc sum of (4x-1)/(2x^2+1)>= -3 = -x-y-z-3 which is true since sum of cyc inequalities: (4x-1)/(2x^2+1)>=-x-1