Let $x,y,z>0$ show that $$(x+y+z)^8\ge 3^7(x^2+y^2+z^2)x^2y^2z^2$$
Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them. Things I have done so far: We know that $$(x^2+y^2+z^2)x^2y^2z^2\le\dfrac{1}{3}(x^2y^2+y^2z^2+x^2z^2)^2$$ it is sufficient prove following $$(x+y+z)^8\ge 3^6(x^2y^2+y^2z^2+z^2x^2)^2$$ or $$(x+y+z)^4\ge 3^3(x^2y^2+y^2z^2+z^2x^2)$$
More way.
By AM-GM $(x+y+z)^3\geq27xyz$ and $3xyz\leq\frac{(xy+xz+yz)^2}{x+y+z}$.
Thus, $$3^7(x^2+y^2+z^2)x^2y^2z^2=27(x^2+y^2+z^2)\cdot27xyz\cdot3xyz\leq$$ $$\leq27(x^2+y^2+z^2)\cdot(x+y+z)^3\cdot\frac{(xy+xz+yz)^2}{x+y+z}=$$ $$=27(x^2+y^2+z^2)(x+y+z)^2(xy+xz+yz)^2$$
and it remains to prove that $$(x+y+z)^6\geq27(x^2+y^2+z^2)(xy+xz+yz)^2.$$ Let $x^2+y^2+z^2=t^3(xy+xz+yz)$.
Hence, we need to prove that $$(t^3+2)^3\geq27t^3$$ or $$t^3+2\geq3t,$$
which is AM-GM again: $$t^3+2=t^3+1+1\geq3\sqrt[3]{t^3\cdot1\cdot1}=3t.$$ Done!
Another way.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that $f(w^3)\geq0$, where $f(w^3)=u^8-(3u^2-2v^2)w^6$, which is decreasing function.
Thus, it remains to prove our inequality for a maximal value of $w^3$.
$x$, $y$ and $z$ are positive roots of the equation $(X-x)(X-y)(X-z)=0$ or
$X^3-3uX^2+3v^2X-w^3=0$ or $w^3=X^3-3uX^2+3v^2X$,
which says that the line $Y=w^3$ and a graph of $Y=X^3-3uX^2+3v^2X$
have three common points and $w^3$ gets a maximal value,
when a line $Y=w^3$ is a tangent line to the graph of $Y=X^3-3uX^2+3v^2X$,
which happens for equality case of two variables. Draw it!
Since our inequality is homogeneous and eighth degree,
it's enough to prove our inequality for $y=z=1$, which gives $$(x-1)^2(x^6+18x^5+147x^4+724x^3+234x^2+1536x+256)\geq0$$ Done!