uvw method i think.i will post soloution.

mmaht wrote: uvw method i think.i will post soloution. yes uvw killed this problem. my soloution:by uvw we must prove that(with some details):$-7w^6u^2-6w^6v^2+18u^3v^2w^3+52uv^4w^3-36u^5w^3-90v^6u^2-12v^8+81u^4v^4\geq0$ .so if we put $\frac{v}{w}=a$ and $\frac{u}{v}=b$ we have $a\geq b\geq 1$ so we must prove that $-7w^8b^2a^2-6w^8a^2+18b^3a^5w^8+52ba^5w^8-36b^5a^5w^8-90a^8w^8b^2-12a^8w^8+81b^4a^8w^8\geq0$ $\rightarrow$ $f(a,b)=-7b^2-6+18b^3a^3+52ba^3-36b^5a^3-90a^6b^2-12a^6+81b^4a^6\geq0$ but $f(a,b)\geq f(b,b)$ and $f(b,b)=-7b^2-6+6b^6+52b^4-126b^8+81b^{10}=(b-1)(b+1)(3b^2+1)(3b^2-2)(9b^4-2b^2-3)\geq 0$ so we are done (if i found an easier way i will post that)

nima1376 wrote: if $x,y,z>0$ are postive real numbers such that $x^{2}+y^{2}+z^{2}=x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}$ prove that \[((x-y)(y-z)(z-x))^{2}\leq 2((x^{2}-y^{2})^{2}+(y^{2}-z^{2})^{2}+(z^{2}-x^{2})^{2})\] Homogenize right side to degree six and WLOG $x\leq y \leq z$, $y = x + u$, $z = x + u + v$, then the inequality becomes \begin{align*} 4 u^8 + 16 u^7 v + 38 u^6 v^2 + 58 u^5 v^3 + 49 u^4 v^4 + 20 u^3 v^5 + 3 u^2 v^6 + \\ (32 u^7 + 112 u^6 v + 236 u^5 v^2 + 310 u^4 v^3 + 216 u^3 v^4 + 70 u^2 v^5 + 8 u v^6) x + \\ (112 u^6 + 336 u^5 v + 629 u^4 v^2 + 698 u^3 v^3 + 389 u^2 v^4 + 96 u v^5 + 8 v^6) x^2 + \\ (224 u^5 + 560 u^4 v + 928 u^3 v^2 + 832 u^2 v^3 + 336 u v^4 + 48 v^5) x^3 + \\ (268 u^4 + 536 u^3 v + 764 u^2 v^2 + 496 u v^3 + 108 v^4) x^4 + \\ (176 u^3 + 264 u^2 v + 312 u v^2 + 112 v^3) x^5 + \\ (48 u^2 + 48 u v + 48 v^2) x^6 \geq 0 \end{align*} which is obvious.

nima1376 wrote: if $x,y,z>0$ are postive real numbers such that $x^{2}+y^{2}+z^{2}=x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}$ prove that \[((x-y)(y-z)(z-x))^{2}\leq 2((x^{2}-y^{2})^{2}+(y^{2}-z^{2})^{2}+(z^{2}-x^{2})^{2})\] I think this inequality not enough hard for a problem in Iran TST 2014. Indeed, we can prove the following sharper inequality holds \[2\big[(x-y)(y-z)(z-x)\big]^{2}\leq (x^{2}-y^{2})^{2}+(y^{2}-z^{2})^{2}+(z^{2}-x^{2})^{2}.\] Without loss of generality, assume that $x \geq y \geq z > 0,$ then \[ \begin{aligned} 2\big[(x-y)(y-z)(z-x)\big]^{2} \ & \leq 2(x-y)^2x^2y^2 \leq 2(x-y)^2(x^2y^2+y^2z^2+z^2x^2) \\ & =2(x-y)^2(x^2+y^2+z^2) \leq 2(x-y)^2(x^2+y^2+2xy) \\ & =2(x^2-y^2)^2 \leq (x^2-y^2)^2+(y^2-z^2)^2+(z^2-x^2)^2.\end{aligned}\] Hence, the proof is completed.

Very nice. Very Powerful.

Expanding out this problem is pretty feasible: Spotting that the right-hand side and left-hand side have different degrees, we seek to use the given condition to homogenize the inequality. In particular, we will prove that the difference \[2\left(x^2y^2 + y^2z^2 + z^2x^2\right)\left(\left(x^2 - y^2\right)^2 + \left(y^2 - z^2\right)^2 + \left(z^2 - x^2\right)^2\right) - \left(x^2 + y^2 + z^2\right)\left((x - y)(y - z)(z - x)\right)^2\] is nonnegative. Upon expansion and cancellation, this reduces to the inequality \[0 \le 3\left(\sum\limits_{\text{sym}} x^6y^2\right) + 2\left(\sum\limits_{\text{sym}} x^5y^3\right) + 2\left(\sum\limits_{\text{cyc}} x^6yz\right) - 2\left(\sum\limits_{\text{sym}} x^5y^2z\right) - 6\left(\sum\limits_{\text{cyc}} x^4y^4\right) - 2\left(\sum\limits_{\text{cyc}} x^3y^2z^2\right).\] This inequality is implied by adding the following inequalities: \begin{align*} 2\left(\sum\limits_{\text{sym}} x^5y^3\right) - 2\left(\sum\limits_{\text{sym}} x^5y^2z\right) = 2\sum\limits_{\text{cyc}} z^5\left(x^3 + y^3 - x^2y - xy^2\right) = 2\sum\limits_{\text{cyc}} z^5(x + y)\left(x - y\right)^2 &\ge 0 \\ 3\left(\sum\limits_{\text{sym}} x^6y^2\right) - 6\left(\sum\limits_{\text{cyc}} x^4y^4\right) = 3\sum\limits_{\text{cyc}} x^6y^2 + x^2y^6 - 2x^4y^4 = 3\sum\limits_{\text{cyc}} x^2y^2\left(x + y\right)^2\left(x - y\right)^2 &\ge 0 \\ 2\left(\sum\limits_{\text{cyc}} x^6yz\right) - 2\left(\sum\limits_{\text{cyc}} x^3y^2z^2\right) = 2xyz\left(x^5 + y^5 + z^5 - x^2y^2z - xy^2z^2 - x^2yz^2\right) &\ge 0 \end{align*} where the last inequality follows from AM-GM/Muirhead: $x^5 + y^5 + z^5 \ge x^2y^2z + xy^2z^2 + x^2yz^2.$ $\square$

If $x,y,z$ are postive real numbers such that $xy+yz+zx=x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}$.Prove that \[3(x-y)^{2}(y-z)^{2}(z-x)^{2}\leq (x^{2}-y^{2})^{2}+(y^{2}-z^{2})^{2}+(z^{2}-x^{2})^{2}\]

The following is also true: If $x,y,z$ are postive real numbers such that $xy+yz+zx=x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}$. Prove that \[4(x-y)^{2}(y-z)^{2}(z-x)^{2}\leq (x^{2}-y^{2})^{2}+(y^{2}-z^{2})^{2}+(z^{2}-x^{2})^{2}\]

The maximal result here is the following. Let $x$, $y$ and $z$ be non-negative numbers such that $xy+yz+zx=x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}$. Prove that : \[12(x-y)^{2}(y-z)^{2}(z-x)^{2}\leq (x^{2}-y^{2})^{2}+(y^{2}-z^{2})^{2}+(z^{2}-x^{2})^{2}\]

Hello! One for the initial problem Cauchy-Schwarz gives $(x^2y^2+y^2z^2+z^2x^2)\left[(x^2-y^2)^2+(y^2-z^2)^2+(z^2-x^2)^2\right]\geq$ $\geq \left(xy|x^2-y^2|+yz|y^2-z^2|+zx|z^2-x^2|\right)^2\geq \left|xy(x^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2)\right|^2=$ $=\left[(x-y)(y-z)(z-x)(x+y+z)\right]^2$ (we also used that $|a|+|b|+|c|\geq |a+b+c| \ \forall a,b,c\in \mathbb{R}$). Thus,it remains to show that $(x^2y^2+y^2z^2+z^2x^2)\left((x-y)(y-z)(z-x)\right)^2\leq 2\left((x-y)(y-z)(z-x)(x+y+z)\right)^2$ This is equivalent to $\left((x-y)(y-z)(z-x)\right)^2\left(2(x+y+z)^2-x^2y^2-y^2z^2-z^2x^2\right)\geq 0$, which,using the given condition,becomes $\left((x-y)(y-z)(z-x)\right)^2(x^2+y^2+z^2+4xy+4yz+4zx)\geq 0$ which is obvious.

When I got the paper ,there was no 2 in RHS ........but I think it is also right

All the sums in the following solutions are cyclic in $x,y,z$ . We will prove the inequality $:=$ $$\left((x-y)(y-z)(z-x)\right)^{2}\leq ((x^{2}-y^{2})^{2}+(y^{2}-z^{2})^{2}+(z^{2}-x^{2})^{2})$$$$\iff ((x-y)(y-z)(z-x))^{2}\left(\sum x^2\right)\leq ((x^{2}-y^{2})^{2}+(y^{2}-z^{2})^{2}+(z^{2}-x^{2})^{2}) \left(\sum x^2y^2\right)$$ Note that $:=$ $$ \left(\sum xy(x^2-y^2) \right)^2 \stackrel {\text {C-S}}{\leq} \text {RHS}$$ Also note that by a simple calculation $$\left((x-y)(y-z)(z-x)(x+y+z)\right)^2 = \left(\sum xy(x^2-y^2) \right)^2 $$ So we have $:=$ $$((x-y)(y-z)(z-x))^{2}\left(\sum x^2\right)\leq \left((x-y)(y-z)(z-x)(x+y+z)\right)^2 = \left( \sum xy(x^2-y^2) \right)^2 \leq ((x^{2}-y^{2})^{2}+(y^{2}-z^{2})^{2}+(z^{2}-x^{2})^{2}) \left(\sum x^2y^2\right) $$ We are done $\blacksquare$

arqady wrote: The maximal result here is the following. Let $x$, $y$ and $z$ be non-negative numbers such that $xy+yz+zx=x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}$. Prove that : \[12(x-y)^{2}(y-z)^{2}(z-x)^{2}\leq (x^{2}-y^{2})^{2}+(y^{2}-z^{2})^{2}+(z^{2}-x^{2})^{2}\] Dear arqady,after a long-time calculation of pqr method,i think the inequality is probably wrong. First i maltiple LHS by $xy+yz+zx$ and RHS by $x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}$ to make the degrees to the same,then i WLOG let $p=1$,and make it to an r's quadralic function:$$(162q-12)r^2+(32q-2-104q^2)r+q^2(5q-1)^2\geq0$$,and find when $\lambda<q<\frac{1}{3}$,where$\lambda\approx0.328$ is the root of the equition $$4050x^5-4624x^4+1946x^3-372x^2+32x-1=0$$, $\Delta>0$ and the axis of the quadralic function can be obtained by $r$ (that's because the axis lies in $x=\frac{(1-\sqrt{1-3q})^2(1+2\sqrt{1-3q})}{27}$ which is the biggest value $r$ can obtain),hence the inequality is wrong.

Attachments:

Just-a-SuperNova wrote: arqady wrote: The maximal result here is the following. Let $x$, $y$ and $z$ be non-negative numbers such that $xy+yz+zx=x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}$. Prove that : \[12(x-y)^{2}(y-z)^{2}(z-x)^{2}\leq (x^{2}-y^{2})^{2}+(y^{2}-z^{2})^{2}+(z^{2}-x^{2})^{2}\] Dear arqady,after a long-time calculation of pqr method,i think the inequality is probably wrong. First i maltiple LHS by $xy+yz+zx$ and RHS by $x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}$ to make the degrees to the same,then i WLOG let $p=1$,and make it to an r's quadralic function:$$(162q-12)r^2+(32q-2-104q^2)r+q^2(5q-1)^2\geq0$$,and find when $\lambda<q<\frac{1}{3}$,where$\lambda\approx0.328$ is the root of the equition $$4050x^5-4624x^4+1946x^3-372x^2+32x-1=0$$, $\Delta>0$ and the axis of the quadralic function can be obtained by $r$ (that's because the axis lies in $x=\frac{(1-\sqrt{1-3q})^2(1+2\sqrt{1-3q})}{27}$ which is the biggest value $r$ can obtain),hence the inequality is wrong. Oh i made a mistake,$r$ should have a minimal $\frac{(1+\sqrt{1-3q})^2(1-2\sqrt{1-3q})}{27}$,and it is bigger than the axis as $q$ lies in $(\lambda,\frac{1}{3})$.And when $r$ is at its maximum or minimal,there must be some two equal in $x,y,z$,and the inequality is apprently true. So we just need to check $(0,\frac{2}{27})$,and its minimal is obtained when $r$ is at its maximum or minimal so the inequality is true. let$x=0,y=\frac{\sqrt{5}-1}{2},z=\frac{\sqrt{5}+1}{2}$ we find 12 is the best.