Notice that its equivalent to $\frac{a}{b+c} + \frac{b}{c+a} +\frac{c}{a+b} \geq \frac{3}{2} $ for $a=2x^2-x+y+z $ , $b=2y^2 +x-y+z$, $c=2z^2+x+y-z$ This can be rewritten as $\sum \frac{(a-b)^2}{(c+a)(c+b)} \geq 0$ So since $a+b , b+c , c+a >0$ it is true and equality holds when $a=b=c \implies 2x^2-x+y+z=2y^2 +x-y+z=2z^2+x+y-z$ $\implies x^2-x=y^2-y=z^2-z \implies (x,y,z)=(t,t,t), (t,t,1-t)$ and permutations

Add $2$ at each fraction and apply AM-HM to finish. Equality holds if $x+y^2+z^2=x^2+y+z^2=x^2+y^2+z,$ and so $x^2-x=y^2-y=z^2-z$, that is $|x-1/2|=|y-1/2|=|z-1/2|$, hence $(x,y,z)=(k,k,1-k)$ or $(k,k,k)$ and permutations.

Batapan wrote: which is Nesbitt's Inequality We should be a bit careful here, since the numbers $a,b,c$ are not necessarily positive.

very good problem $\sum{\frac{2x^2-x+y+z}{x+y^2+z^2}}\geq 3 (?)$ $\sum{\frac{2x^2-x+y+z}{x+y^2+z^2}}+2\geq 9 (?)$ $\sum{\frac{2(x^2+y^2+z^2)x+y+z}{x+y^2+z^2}}\geq 9 (?)$ $(2(x^2+y^2+z^2)x+y+z)\sum{\frac{1}{x+y^2+z^2}}\geq 9 (?)$ By $TITU$ $(2(x^2+y^2+z^2)x+y+z)\sum{\frac{1}{x+y^2+z^2}}\geq (2(x^2+y^2+z^2)x+y+z)\sum{\frac{9}{2(x^2+y^2+z^2)+x+y+z}}=9$ and we are done

@Orestis_Lignos When other questions will be shared in aops?

Prove that for all non-negative real numbers $x,y,z$, not all equal to $0$, the following inequality holds $$\dfrac{x+y^2+z^2}{2x^2-x+y+z}+\dfrac{x^2+y+z^2}{2y^2+x-y+z}+\dfrac{x^2+y^2+z}{2z^2+x+y-z} \geq 3$$

Orestis_Lignos wrote: Prove that for all non-negative real numbers $x,y,z$, not all equal to $0$, the following inequality holds $\displaystyle \dfrac{2x^2-x+y+z}{x+y^2+z^2}+\dfrac{2y^2+x-y+z}{x^2+y+z^2}+\dfrac{2z^2+x+y-z}{x^2+y^2+z}\geq 3.$ Determine all the triples $(x,y,z)$ for which the equality holds. $$\iff \dfrac{2x^2-x+y+z}{x+y^2+z^2}+2+\dfrac{2y^2+x-y+z}{x^2+y+z^2}+2+\dfrac{2z^2+x+y-z}{x^2+y^2+z}+2\geq 9$$$$\iff(x+y^2+z^2+x^2+y+z^2+x^2+y^2+z)(\dfrac{1}{x+y^2+z^2}+\dfrac{1}{x^2+y+z^2}+\dfrac{1}{x^2+y^2+z})\geq 9$$which is direct from the $$(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})\geq 9$$Where $ a=x+y^2+z^2>0, b=x^2+y+z^2>0, c=x^2+y^2+z>0.$ Equality for $x+y^2+z^2=x^2+y+z^2=x^2+y^2+z$ The above equality is equivalent to $(x-\dfrac{1}{2})^2=(y-\dfrac{1}{2})^2=(z-\dfrac{1}{2})^2,$ from which we analytically deduce all cases of equality, which are:$1) x=y=z,2) x=y=1-z,3) x=1-y=z,4) 1-x=y=z.$ here

My solution is also simmilar to sqing $\displaystyle \dfrac{2x^2-x+y+z}{x+y^2+z^2}+\dfrac{2y^2+x-y+z}{x^2+y+z^2}+\dfrac{2z^2+x+y-z}{x^2+y^2+z}=\sum{\frac{2x^2-x+y+z}{x+y^2+z^2}}$ $\sum{\frac{2x^2-x+y+z}{x+y^2+z^2}}\geq 3$ $\sum{\frac{2x^2-x+y+z}{x+y^2+z^2}}+2\geq 3+6$ $\sum{\frac{2x^2+2y^2+2z^2+x+y+z}{x+y^2+z^2}}\geq 9$ $\dfrac{2x^2+2y^2+2z^2+x+y+z}{x+y^2+z^2}+\dfrac{2x^2+2y^2+2z^2+x+y+z}{x^2+y+z^2}+\dfrac{2x^2+2y^2+2z^2+x+y+z}{x^2+y^2+z}\geq 9$ $(2x^2+2y^2+2z^2+x+y+z)(\dfrac{1}{x+y^2+z^2}+\dfrac{1}{x^2+y+z^2}+\dfrac{1}{x^2+y^2+z})\geq 9$ $(x+y^2+z^2+x^2+y+z^2+x^2+y^2+z)(\dfrac{1}{x+y^2+z^2}+\dfrac{1}{x^2+y+z^2}+\dfrac{1}{x^2+y^2+z})\geq 9$ Let $a=x+y^2+z^2$ , $b=x^2+y+z^2$ , $c=x^2+y^2+z$ $=>$ $(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})\geq 9$ Which can easly be proven by $AM-GM$ $a+b+c\geq3\sqrt[3]{abc}$ , $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\geq 3\sqrt[3]{\dfrac{1}{abc}}$ $(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})$$\geq 3\sqrt[3]{abc}$ $3\sqrt[3]{\dfrac{1}{abc}}$ $(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})\geq$3x3$\sqrt[3]{abc}$ x $\sqrt[3]{\dfrac{1}{abc}}$ $(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})\geq$9$\sqrt[3]{abc \dfrac{1}{abc}}$ $(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})\geq$9$\sqrt[3]{1}$ $(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})\geq9$

sqing wrote: Prove that for all non-negative real numbers $x,y,z$, not all equal to $0$, the following inequality holds $$\dfrac{x+y^2+z^2}{2x^2-x+y+z}+\dfrac{x^2+y+z^2}{2y^2+x-y+z}+\dfrac{x^2+y^2+z}{2z^2+x+y-z} \geq 3$$ very easy $\sum{\frac{x+y^2+z^2}{2x^2-x+y+z}} \geq 3(?)$ $2\sum{\frac{x+y^2+z^2}{2x^2-x+y+z}}\geq 6(?)$ $\sum{\frac{2(y^2+z^2)+2x}{2x^2-x+y+z}}+1 \geq 9(?)$ $\sum{\frac{2(x^2+y^2+z^2)+x+y+z}{2x^2-x+y+z}} \geq 9(?)$ By $TITU$ $(2(x^2+y^2+z^2)+x+y+z)\sum{\frac{1}{2x^2-x+y+z}} \geq 9$ and we are done

Alternative. Add 1 to each term: \[ \frac{2x^2-x+y+z}{x+y^2+z^2}+1 = \frac{2x^2+y^2+y+z^2+z}{x+y^2+z^2}. \]Now set $a=x+y^2+z^2$, $b=x^2+y+z^2$ and $c=x^2+y^2+z$. Observe that $2x^2+y^2+y+z^2+z = b+c$, so it suffices to prove \[ \sum \frac{b+c}{a}\ge 6, \]which is trivial ($b/a+a/b\ge 2$ by AM-GM and so on).

Unfortunately, we seem to have no perfect solution yet.

Let $A=x^2-x$, $B=y^2-y$, $C=z^2-z$, $a=x+y^2+z^2$, $b=x^2+y+z^2$, $c=x^2+y^2+z$. Then, $$\dfrac{2x^2-x+y+z}{x+y^2+z^2}+\dfrac{2y^2+x-y+z}{x^2+y+z^2}+\dfrac{2z^2+x+y-z}{x^2+y^2+z}\geq 3\iff \sum_{cyc}\dfrac{2A-B-C}{a} \geq 0 \iff \sum_{cyc} \dfrac{(A-B)^2}{ab} \geq 0$$(In the last part, I used $b-a=A-B$, etc.) Which is true, since $ab$, $bc$, $ac$ are all non-negative. Equality occurs if and only if $A=B=C$ i.e $x^2-x=y^2-y=z^2-z$.

We need to prove $$\sum_{cyc}\frac{2x^2-x+y+z}{x+y^2+z^2}\geq3\iff\sum_{cyc}\frac{2(x^2+y^2+z^2)+x+y+z}{x+y^2+z^2}\geq9\iff (2(x^2+y^2+z^2)+x+y+z)\sum_{cyc}\frac{1}{x+y^2+z^2}\geq9$$From Titu's Lemma we have that $$(2(x^2+y^2+z^2)+x+y+z)\sum_{cyc}\frac{1}{x+y^2+z^2}\geq(2(x^2+y^2+z^2)+x+y+z)\frac{9}{2(x^2+y^2+z^2)+x+y+z}=9$$Now we need to look at when equality holds. $$\frac{1}{x+y^2+z^2}=\frac{1}{x^2+y+z^2}=\frac{1}{x^2+y^2+z}\iff x^2-x=y^2-y=z^2-z$$$x^2-x=y^2-y\iff(x-y)(x+y-1)=0\implies y=x$ or $y=1-x$ Solving $x^2-x=z^2-z$ and $y^2-y=z^2-z$ analogously we come to the conclusion that equality holds for: $$(x,y,z)=(a,a,a) \forall a>0$$$$(x,y,z)=(a,a,1-a) \forall a\in[0,1]$$$$(x,y,z)=(a,1-a,a) \forall a\in[0,1]$$$$(x,y,z)=(1-a,a,a) \forall a\in[0,1]$$

dancho wrote: We need to prove $$\sum_{cyc}\frac{2x^2-x+y+z}{x+y^2+z^2}\geq3\iff\sum_{cyc}\frac{2(x^2+y^2+z^2)+x+y+z}{x+y^2+z^2}\geq9\iff (2(x^2+y^2+z^2)+x+y+z)\sum_{cyc}\frac{1}{x+y^2+z^2}\geq9$$From Titu's Lemma we have that $$(2(x^2+y^2+z^2)+x+y+z)\sum_{cyc}\frac{1}{x+y^2+z^2}\geq(2(x^2+y^2+z^2)+x+y+z)\frac{9}{2(x^2+y^2+z^2)+x+y+z}=9$$Now we need to look at when equality holds. $$\frac{1}{x+y^2+z^2}=\frac{1}{x^2+y+z^2}=\frac{1}{x^2+y^2+z}\iff x^2-x=y^2-y=z^2-z$$$x^2-x=y^2-y\iff(x-y)(x+y-1)=0\implies y=x$ or $y=1-x$ Solving $x^2-x=z^2-z$ and $y^2-y=z^2-z$ analogously we come to the conclusion that equality holds for: $$(x,y,z)=(a,a,a) \forall a>0$$$$(x,y,z)=(a,a,1-a) \forall a\in[0,1]$$$$(x,y,z)=(a,1-a,a) \forall a\in[0,1]$$$$(x,y,z)=(1-a,a,a) \forall a\in[0,1]$$ dancho, Almost Perfect ! I grade your solution by 9 + (1/9) points / 10 points. Writing $(x,y,z)=(b,b,1-b) \forall b\in[0,1]$ etc. would give full marks ! This problem seems to have slight difficulty with showing equality case.

Kunihiko_Chikaya wrote: dancho wrote: We need to prove $$\sum_{cyc}\frac{2x^2-x+y+z}{x+y^2+z^2}\geq3\iff\sum_{cyc}\frac{2(x^2+y^2+z^2)+x+y+z}{x+y^2+z^2}\geq9\iff (2(x^2+y^2+z^2)+x+y+z)\sum_{cyc}\frac{1}{x+y^2+z^2}\geq9$$From Titu's Lemma we have that $$(2(x^2+y^2+z^2)+x+y+z)\sum_{cyc}\frac{1}{x+y^2+z^2}\geq(2(x^2+y^2+z^2)+x+y+z)\frac{9}{2(x^2+y^2+z^2)+x+y+z}=9$$Now we need to look at when equality holds. $$\frac{1}{x+y^2+z^2}=\frac{1}{x^2+y+z^2}=\frac{1}{x^2+y^2+z}\iff x^2-x=y^2-y=z^2-z$$$x^2-x=y^2-y\iff(x-y)(x+y-1)=0\implies y=x$ or $y=1-x$ Solving $x^2-x=z^2-z$ and $y^2-y=z^2-z$ analogously we come to the conclusion that equality holds for: $$(x,y,z)=(a,a,a) \forall a>0$$$$(x,y,z)=(a,a,1-a) \forall a\in[0,1]$$$$(x,y,z)=(a,1-a,a) \forall a\in[0,1]$$$$(x,y,z)=(1-a,a,a) \forall a\in[0,1]$$ dancho, Almost Perfect ! I grade your solution by 9 + (1/9) points / 10 points. Writing $(x,y,z)=(b,b,1-b) \forall b\in[0,1]$ etc. would give full marks ! This problem seems to have slight difficulty with showing equality case. Thanks, I actually did get full marks on the competition itself and I didn't think it would matter if I use the same variable in the equality cases here.

Orestis_Lignos wrote: Prove that for all non-negative real numbers $x,y,z$, not all equal to $0$, the following inequality holds $\displaystyle \dfrac{2x^2-x+y+z}{x+y^2+z^2}+\dfrac{2y^2+x-y+z}{x^2+y+z^2}+\dfrac{2z^2+x+y-z}{x^2+y^2+z}\geq 3.$ Determine all the triples $(x,y,z)$ for which the equality holds. Milan Mitreski, Serbia $\frac{2x^2-x+y+z}{x+y^2+z^2}+\frac{2y^2-y+z+x}{y+z^2+z^2}+\frac{2z^2-z+x+y}{z+x^2+y^2}$ $=\sum{\frac{x^2+y^2+z}{x+y^2+z^2}}+\sum{\frac{x^2-x+y-y^2}{x+y^2+z^2}}\geq 3+\sum{\frac{x^2-x+y-y^2}{x+y^2+z^2}}\geq 3$ That implies $\sum{\frac{x^2-x+y-y^2}{x+y^2+z^2}}\geq 0$ Add both sides 3 $\sum{\frac{x^2-x+y-y^2}{x+y^2+z^2}+1}=\boxed{\frac{x^2+z^2+y}{x+y^2+z^2}+\frac{x^2+y^2+z}{y+x^2+z^2}+\frac{y^2+z^2+x}{z+x^2+y^2}\geq 3}$. Which is true. The equality case should occur when $x^2+y=x+y^2$, $z^2+y=y^2+z$ $x(x-1)=y(y-1)=z(z-1)$ There are two possibilities and their permutations: $i)$ $\boxed{(x,y,z)=(t,t,t)}$ $ii)$ $\boxed{(x,y,z)=(1-t,t,t)}$ , $b\in [0,1]$ and their permutations. Since $(1-t,t,t)=(1-t,1-t,t)$, the $ii)$ mentions all. $Q.E.D.$

Version 1 Let $x,y,z,u$ be nonnegative reals (Not all equal to $0$). Then prove that $$\sum_{cyc}{\dfrac{3x^2-2x+y+z+u}{x+y^2+z^2+u^2}}\geq 4$$ and determine when does the equality holds?

Generalization 1 Let $x_{1},x_{2},\cdots,n$ ($n\geq 2$) be nonnegative reals (Not all equal to $0$). Then prove that $$\sum_{cyc}{\dfrac{(n-1)x_{1}^2-(n-2)x_{1}+\sum_{2\leq i \leq n}{x_{i}}}{x_{1}+\sum_{2\leq i\leq n}{x_{i}^2}}}\geq n$$ and determine when does the equality holds?

We do a bit of trolling. Subtract one from each term and rearrange to get \[2\sum_{cyc}\frac{z(z-1)}{x^2+y^2+z}\geq \sum_{cyc}\frac{x(x-1)+y(y-1)}{x^2+y^2+z}.\]This follows by rearrangement on \[\left(z(z-1),z(z-1),y(y-1),y(y-1),y(y-1),x(x-1),x(x-1)\right)\]and \[\left(\frac{1}{x^2+y^2+z}, \frac{1}{x^2+y^2+z}, \frac{1}{x^2+y+z^2}, \frac{1}{x^2+y+z^2}, \frac{1}{x+y^2+z^2}, \frac{1}{x+y^2+z^2}\right)\]as $z(z-1)\geq y(y-1) \iff \frac{1}{x^2+y^2+z}\geq \frac{1}{x^2+y+z^2}$.

$Let's say;$ $2x^2-x+y+z=a$ $2y^2+x-y+z=b$ $2z^2+x+y-z=c$ $Then we can see it;$ $a+b=2x^2+2y^2+2z=2*(x^2+y^2+z) b+c=2y^2+2z^2+2y=2*(x^2+z^2+y)$ Now we must prove; $2*a/(b+c)+2b/(a+c)+2c/(a+b)>=3$ From Nesbit; $ a/(b+c)+b/(a+c)+c/(a+b)>=3/2$ So, 2*3\2=3>=3 So, we proved. And also the another question is when does this equality holds. We use Nesbit and Nesbit holds for a=b=c. (x,y,z)=(a,a,1-a),(a,a,a)

The given inequality is equivalent to $\sum_{cyc}{\frac{2x^2+2y^2+2z^2}{x+y^2+z^2}} \geq 9$ (add $2$ to every fraction) which is obviously true by AM-HM. Now equality holds iff $x^2-x=y^2-y=z^-z$. Thus the only solutions are $(p,p,1-p)$, $(q,q,q)$, and $(r,1-r,1-r)$ and its permutations. Note we need to have $p \in [0,1]$, $r \in [0,1]$ and $q > 0$ to satisfy the problem's condition.

Add $3$ to both sides of the inequality to obtain: \[\frac{(x^2+y+z^2)+(x^2+y+z^2)}{x+y^2+z^2} + \frac{(x+y^2+z^2)+(x^2+y+z^2)}{x^2+y+z^2} + \frac{(x+y^2+z^2)+(x^2+y+z^2)}{x^2+y+z^2} \ge 6.\]Now, the solution is clear; let $a = x+y^2+z^2$, $b = x^2+y+z^2$, and $c=x^2+y^2+z$. We wish to show that \[\sum_{sym}\frac{a}{b} \ge 6.\]But since $a,b,c>0$, this follows directly from Muirhead, and we are done. Equality holds iff $a=b=c$, or when $x^2-x=y^2-y = z^2-z$, which implies either $(x,y,z) = (u,u,u), (v,v,1-v)$ and permutations, for any $u>0$ and $v\in[0,1]$. $\blacksquare$

This was A3 at last year's shortlist.

Looks scarry but its just the nesbitt inequality where a,b and c are the numbers on the numerators.

Add $2$ to each fraction. We want to prove that $$(2(x^2+y^2+z^2)+(x+y+z))\sum_{cyc} \frac{1}{x+y^2+z^2} \geq 9$$by Titu´s $$\sum_{cyc} \frac{1}{x+y^2+z^2} \geq \frac{9}{2(x^2+y^2+z^2)+(x+y+z)}$$so the result follows. $\blacksquare$

Immediately we notice that the numerator of each fraction plus twice the denominator is the constant $2(x^2+y^2+z^2)+x+y+z=S$. Therefore we notice this just rearranges into $\sum \frac{a}{S-a} \geq \frac32$ after making the suitable substitutions which follows by Jensens.