$$(zy+xz-yz)(yz+xy-xz)$\ge 2$x^2y^2z^2$ then replace $x^2$ by $1-y^2-z^2$ and put $yz=a$ and $y-z=b$ and we have to prove that $(2a+b^2)(2a^2+b^2)\ge (a^2+b^2) which is true bu C-S since $a\ge\frac{1}{4}

Nice inequality for an olympiad.

Manar-Manth wrote: $$(zy+xz-yz)(yz+xy-xz)$\ge 2$x^2y^2z^2$ then replace $x^2$ by $1-y^2-z^2$ and put $yz=a$ and $y-z=b$ and we have to prove that $(2a+b^2)(2a^2+b^2)\ge (a^2+b^2) which is true bu C-S since $a\ge\frac{1}{4} Don't understand

Manar-Manth wrote: $(zy+xz-yz)(yz+xy-xz) \ge 2 x^2y^2 z^2$ then replace $x^2$ by $1 -y^2- z^2$ and put $yz=a$ and $y-z=b$ and we have to prove that $(2a+b^2)(2a^2+b^2) \ge (a^2+b^2) $ which is true bu C-S since $a \ge \frac{1}{4}$

Is above what you meant?

yes

Manar-Manth wrote: $$(zy+xz-yz)(yz+xy-xz)$\ge 2$x^2y^2z^2$ then replace $x^2$ by $1-y^2-z^2$ and put $yz=a$ and $y-z=b$ and we have to prove that $(2a+b^2)(2a^2+b^2)\ge (a^2+b^2) which is true bu C-S since $a\ge\frac{1}{4} C-S？

http://matol.kz/comments/2926/show

mudok wrote: http://matol.kz/comments/2926/show In that link, the first solution is official solution.

mudok wrote: mudok wrote: http://matol.kz/comments/2926/show In that link, the first solution is official solution. But the second is nicer.

Please, check my solution. We want to prove $$\frac{1}{x^2} \geq 2+(\frac{1}{y}-\frac{1}{z})^2$$ $\frac{1}{4} \leq yz \leq \frac{1-x^2}{2}<\frac{1}{2}$ $x^2=1-y^2-z^2 \leq 1-2yz$ $\frac{1}{x^2}+\frac{1}{2yz} \geq \frac{1}{1-2yz}+\frac{1}{2yz} \geq 4 \to \frac{1}{x^2} \geq 4-\frac{1}{2yz}$ $2-\frac{1}{2yz} \geq \frac{1}{y^2}+\frac{1}{z^2}-\frac{2}{yz}$ $$4y^2z^2+3yz \geq 2y^2+2z^2$$ $(4yz-1)(2yz-1)\leq 0 \to 6yz \geq 8y^2z^2+1$ and$(4y^2-1)(4z^2-1) \geq 0 \to 16y^2z^2+1 \geq 4y^2+4z^2 $ So $8y^2z^2+6yz \geq 16y^2z^2+1 \geq 4y^2+4z^2$ - Q.E.D

The following inequality is also true

The following inequality is also true?

matol.kz wrote: For positive reals $x,y,z\ge \frac{1}{2}$ with $x^2+y^2+z^2=1$, prove this inequality holds $$(\frac{1}{x}+\frac{1}{y}-\frac{1}{z})(\frac{1}{x}-\frac{1}{y}+\frac{1}{z})\ge 2$$ mudok wrote: http://matol.kz/comments/2926/show $(\frac{1}{y^2}-\frac{1}{4})(\frac{1}{z^2}-\frac{1}{4})\ge 0\iff (y-z)^2\le \frac{1}{4}(4yz-1)^2$ $yz\ge \frac{1}{4}, x^2=1-y^2-z^2\le 1-2yz$ $\frac{1}{x^2}-\left(\frac{1}{y}-\frac{1}{z}\right)^2-2=\frac{1}{x^2}-\frac{\left(y-z\right)^2}{y^2z^2}-2\ge\frac{1}{1-2yz}-\frac{\left(4yz-1\right)^2}{4y^2z^2}-2=\frac{\left(4yz-1\right)\left(3y^2z^2+(3yz-1)^2\right)}{4y^2z^2(1-2yz)} \ge 0$ $(\frac{1}{x}+\frac{1}{y}-\frac{1}{z})(\frac{1}{x}-\frac{1}{y}+\frac{1}{z})\ge 2$