Easy enough. If $x^2-yz$, $y^2-zx$, $z^2-xy$ are all positive, we have, by AM-GM, $$(x^2-yz)(y^2-zx)(z^2-xy) \le (\frac{x^2+y^2+z^2-xy-yz-zx}{3})^3 \le (\frac{1}{2})^3 = \frac{1}{8}$$where $x^2+y^2+z^2 - xy-yz-zx \le \frac{3}{2}$ since $(x+y+z)^2 = 1+2(xy+yz+zx) \ge 0 \implies xy+yz+zx \ge -\frac{1}{2}$ If $x^2-yz>0$ and $y^2-zx<0$, $z^2-xy<0$, by AM-GM again, $$(x^2-yz)(y^2-zx)(z^2-xy) = (x^2-yz)(zx-y^2)(xy-z^2) \le (\frac{x^2+xy+xz-yz-y^2-z^2}{3})^3 \le \frac{1}{8}$$since $x^2+xy+xz-yz-y^2-z^2 =x^2+xy+xz+yz-(y+z)^2 \le x^2+xy+xz+yz \le \frac{3}{2}(x^2+y^2+z^2) = \frac{3}{2}$. The final inequality is just $\frac{1}{2}(y-z)^2 \ge 0$ and $(x-2y)^2 + (x-2z)^2 \ge 0$. Therefore, we have $$(x^2-yz)(y^2-zx)(z^2-xy) \le \frac{1}{8}$$Equality does hold at $x=\frac{\sqrt{2}}{2}$, $y=-\frac{\sqrt{2}}{2}$, and $z=0$. $\blacksquare$

$$(x^2+y^2+z^2)^3-8(x^2-yz)(y^2-zx)(z^2-xy) = \frac{1}{2}(x+y+z)^2\sum_{cyc}((y-z)^4+2x^2(y-z)^2+2y^2z^2)\geq 0$$Equality holds on every $x+y+z=0, x^2+y^2+z^2=1$. Edit: 300th post.

rkm0959 wrote: If $x,y,z$ satisfies $x^2+y^2+z^2=1$, find the maximum possible value of $$(x^2-yz)(y^2-zx)(z^2-xy)$$

Let $L=(x^2-yz)(y^2-xz)(z^2-xy)$. First case is when all three of $x^2-yz,y^2-xz,z^2-xy$ are positive. We have $L\le (\frac{x^2+y^2+z^2-xy-xz-yz}{3})^3$. Since $(x+y+z)^2\ge 0$ $\implies $ $(x+y+z)^2-3=2xy+2xz+2yz-2x^2-2y^2-2z^2\ge -3\implies $ $x^2+y^2+z^2-xy-xz-yz\le \frac{3}{2}$. So $L\le \frac{1}{8}$. Second case is when some of them are negative. WLOG let $yz\ge x^2,xz\ge y^2$. We have $L\le (\frac{xz+yz-xy+z^2-y^2-x^2}{3})^3$. Since $(2x-z)^2+(2y-z)^2+2(x+y)^2+4x^2+4y^2\ge 0$ $\implies $ $x^2+y^2-z^2+xy-xz-yz\ge -\frac{3}{2}$$\implies $ $L\le \frac{1}{8}$.

rkm0959 wrote: If $x,y,z$ satisfies $x^2+y^2+z^2=1$, find the maximum possible value of $$(x^2-yz)(y^2-zx)(z^2-xy)$$ Let $\prod_{cyc}(x^2-yz)\leq k$ or $\prod_{cyc}(x^2-yz)\leq k(x^2+y^2+z^2)^3$. Since the last inequality is homogeneous, we can assume $x=y=1$ and $z=-2$ and we obtain $k\geq\frac{1}{8}$. Thus, it remains to prove that $(x^2+y^2+z^2)^3\geq8\prod_{cyc}(x^2-yz)$. Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$. Hence, $\prod_{cyc}(x^2-yz)=\sum_{cyc}(x^4yz-x^3y^3)=27u^3w^3-27uv^3w^3+3w^6-(27v^6-27uv^2w^3+3w^6)=27u^3w^3-27v^6$. Id est, our inequality is a linear inequality of $w^3$. Thus, it remains to prove our inequality for extremal value of $w^3$, which happens for equality case of two variables. Since our inequality is homogeneous and even degree, we can assume $y=z=1$, which gives $(x+2)^2((x^2-2x+2)^2+2x^2)\geq0$, which is obvious. Hence, the answer is $\frac{1}{8}$.

rkm0959 wrote: If $x,y,z$ satisfies $x^2+y^2+z^2=1$, find the maximum possible value of $$(x^2-yz)(y^2-zx)(z^2-xy)$$ We need to prove \[8(x^2-yz)(y^2-zx)(z^2-xy) \leqslant (x^2+y^2+z^2)^3,\]Suppose $(x^2-yz)(y^2-zx) \geqslant 0,$ then If $z^2-xy \leqslant 0$ then \[8(x^2-yz)(y^2-zx)(z^2-xy) \leqslant 0 \leqslant (x^2+y^2+z^2)^3.\]If $z^2-xy \geqslant 0,$ apply AM-GM Inequality, we have \[4(x^2-yz)(y^2-zx) \leqslant (x^2-yz+y^2-zx)^2.\]Hence, we need to prove that \[(x^2+y^2+z^2)^3 \geqslant 2(x^2-yz+y^2-zx)^2(z^2-xy),\]equivalent to \[\left [2(x^2+y^2)(z^2-xy)+(x^2+y^2+z^2)(x+y-z)^2 \right ](x+y+z)^2 \geqslant 0.\]Which is true. The proof is completed.

rkm0959 wrote: If $x,y,z$ satisfies $x^2+y^2+z^2=1$, find the maximum possible value of $$(x^2-yz)(y^2-zx)(z^2-xy)$$ If $a,b,c$ be positive real numbers such that $a^{2}>bc$ . Then $$(a^{2}-bc)^{2}>(b^{2}-ca)(c^{2}-ab)$$Japanese MO 2003, Problem 3

rkm0959 wrote: If $x,y,z$ satisfies $x^2+y^2+z^2=1$, find the maximum possible value of $$(x^2-yz)(y^2-zx)(z^2-xy)$$ Given $a,b,c \in R$. Prove that we have the inequality: $$(a^2-bc)(b^2-ca)(c^2-ab) \le \frac{(a^2+b^2+c^2)^3}{8}$$Let $a,b,c$ be reals such that $a+b+c=1.$ Prove that $$(ab-c^2)(bc-a^2)(ca-b^2)\le \frac{11 + 5\sqrt{5}}{54}$$Positive numbers $a,b,c$ are such that $a^2+b^2+c^2=2abc+1$.Find the maximum of $(a^2-bc)(b^2-ca)(c^2-ab)$

rkm0959 wrote: If $x,y,z$ satisfies $x^2+y^2+z^2=1$, find the maximum possible value of $$(x^2-yz)(y^2-zx)(z^2-xy)$$ Let $a=x^{2}-yz$, b, c respectively $$a^{2}+b^{2}+c^{2}+2abc=(x^2+y^2+z^2)^3-(x+y+z)^2(x^2y^2+y^2z^2+z^2x^2) \le 1$$ trivial by AM GM

rkm0959 wrote: Easy enough. If $x^2-yz$, $y^2-zx$, $z^2-xy$ are all positive, we have, by AM-GM, $$(x^2-yz)(y^2-zx)(z^2-xy) \le (\frac{x^2+y^2+z^2-xy-yz-zx}{3})^3 \le (\frac{1}{2})^3 = \frac{1}{8}$$where $x^2+y^2+z^2 - xy-yz-zx \le \frac{3}{2}$ since $(x+y+z)^2 = 1+2(xy+yz+zx) \ge 0 \implies xy+yz+zx \ge -\frac{1}{2}$ If $x^2-yz>0$ and $y^2-zx<0$, $z^2-xy<0$, by AM-GM again, $$(x^2-yz)(y^2-zx)(z^2-xy) = (x^2-yz)(zx-y^2)(xy-z^2) \le (\frac{x^2+xy+xz-yz-y^2-z^2}{3})^3 \le \frac{1}{8}$$since $x^2+xy+xz-yz-y^2-z^2 =x^2+xy+xz+yz-(y+z)^2 \le x^2+xy+xz+yz \le \frac{3}{2}(x^2+y^2+z^2) = \frac{3}{2}$. The final inequality is just $\frac{1}{2}(y-z)^2 \ge 0$ and $(x-2y)^2 + (x-2z)^2 \ge 0$. Therefore, we have $$(x^2-yz)(y^2-zx)(z^2-xy) \le \frac{1}{8}$$Equality does hold at $x=\frac{\sqrt{2}}{2}$, $y=-\frac{\sqrt{2}}{2}$, and $z=0$. $\blacksquare$ Why x^2+xy+xz+yz <=(3/2)(x^2+y^2+z^2)?

IsoLyS wrote: $$(x^2+y^2+z^2)^3-8(x^2-yz)(y^2-zx)(z^2-xy) = \frac{1}{2}(x+y+z)^2\sum_{cyc}((y-z)^4+2x^2(y-z)^2+2y^2z^2)\geq 0$$Equality holds on every $x+y+z=0, x^2+y^2+z^2=1$. Edit: 300th post. Any particular motivation for this ?

rkm0959 wrote: If $x,y,z$ satisfies $x^2+y^2+z^2=1$, find the maximum possible value of $$(x^2-yz)(y^2-zx)(z^2-xy)$$ Can someone help in the last expression? We have prove that the maximum is $\frac18$ Let $p=x+y+z,q=xy+yz+zx,r=xyz$ Then, the LHS expression equals $$p^3r-q^3$$We have to prove $$8p^3r-8q^3\leq (p^2-2q)^3$$$$RHS-LHS=p^2(p^4-6p^2q+12q^2-8pr)$$How do I prove that this expression is always positive?

Math-wiz wrote: $$RHS-LHS=p^2(p^4-6p^2q+12q^2-8pr)$$How do I prove that this expression is always positive? $=p^2\left(\left(p^2-3q\right)^2+2\left(q^2-3pr\right)+\left(q^2-2pr\right)\right){\color{red}{\ge}}\;0$

Victoria_Discalceata1 wrote: Math-wiz wrote: $$RHS-LHS=p^2(p^4-6p^2q+12q^2-8pr)$$How do I prove that this expression is always positive? $=p^2\left(\left(p^2-3q\right)^2+2\left(q^2-3pr\right)+\left(q^2-2pr\right)\right){\color{red}{\ge}}\;0$ Thanks a lot

Sir, how is $q^2-2pr\geq 0$? It is not true for example when $q=-154, p=1, r=-14*8*7$, which is solvable to get $8,7,-14$

It is true all right because $q^2-2pr=\sum x^2y^2$. You have your signs messed up.

Ohk. My bad. Thnaks a lot for the solution:)

rkm0959 wrote: If $x,y,z$ satisfies $x^2+y^2+z^2=1$, find the maximum possible value of $$(x^2-yz)(y^2-zx)(z^2-xy)$$ We need to prove:$$8(x^2-yz)(y^2-zx)(z^2-xy) \leq (x^2+y^2+z^2)^3$$Assume $(x^2 -yz)(y^2 -zx) \geq 0$. Then: $RHS - LHS = \frac{(x^2 -xy+y^2)(\sum\limits_{cyc} x)^2(\sum\limits_{cyc}x^2)^2+2(x^2-yz)(y^2-zx)(\sum\limits_{cyc} x)^2 (x+y-z)^2}{x^2+xy+y^2} \geq 0$

100th post The answer is $\frac{1}{8}$, equality holds when $(x,y,z)=(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2},0)$ To see this is the maximum, notice that when all of the variables have the same sign then the expression is negative. Therefore we may WLOG assume $x\geq y\geq 0\geq z$. Replace $z$ by $-z$ then it suffices to show $$(x^2+yz)(y^2+zx)(z^2-xy)\leq \frac{1}{8}$$for every $x,y,z\geq 0$. Now we need $z^2>xy$, so by AM-GM, we have $$(x^2+yz)(y^2+zx)(z^2-xy)\leq \left(\frac{x^2+y^2+z^2+yz+zx-xy}{3}\right)^3$$Hence it suffices to show $$2(yz+zx-xy)\leq 1=x^2+y^2+z^2$$This rearranges to $(x+y-z)^2\geq 0$, which is obvious.

Let $a,b,c$ be reals such that $a^2+b^2+c^2=1.$ Prove that $$(a^2-bc)(b^2-kca)(c^2-ab)\le \frac{k}{8}$$Where $k\geq 1.$