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Hard and excellent. $Lemma:$ $\frac{a^z}{z}+\frac{b^x}{x}+\frac{c^y}{y} \ge abc$. $Proof:$ This is some weighted stuff, since $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$, $\frac{a^z}{z}+\frac{b^x}{x}+\frac{c^y}{y} \ge (a^z)^{\frac{1}{z}}(b^x)^{\frac{1}{x}}(c^y)^{\frac{1}{y}}=abc$, if nothing just $ln$ both sides and use the concavity. Now by $AM-GM$ (everything there stays positive) we get: $LHS \ge 3\sqrt[3]{\frac{(x^2+y^2)(y^2+z^2)(z^2+x^2)*\frac{1}{\frac{1}{z}-\frac{a^z}{z}}*\frac{1}{\frac{1}{x}-\frac{b^x}{z}}*\frac{1}{\frac{1}{y}-\frac{c^y}{y}}}{xyz}} \ge 3\sqrt[3]{\frac{(x^2+y^2)(y^2+z^2)(z^2+x^2)}{xyz}}* \sqrt[3]{\frac{1}{\frac{1}{\frac{1}{z}-\frac{a^z}{z}}*\frac{1}{\frac{1}{x}-\frac{b^x}{x}}*\frac{1}{\frac{1}{y}-\frac{c^y}{y}}}} \ge 9\sqrt[3]{\frac{(x^2+y^2)(y^2+z^2)(z^2+x^2)}{xyz}} \frac{1}{\frac{1}{z}-\frac{a^z}{z}+\frac{1}{x}-\frac{b^x}{x}+\frac{1}{y}-\frac{c^y}{y}} \ge 9\sqrt[3]{\frac{(x^2+y^2)(y^2+z^2)(z^2+x^2)}{xyz}}* \frac{1}{1-abc}$ by the Lemma, so we are left proving $3\sqrt[3]{\frac{(x^2+y^2)(y^2+z^2)(z^2+x^2)}{xyz}} \ge 2(x+y+z)$, that is fairly straightforward equivalent to $27(x^2+y^2)(y^2+z^2)(z^2+x^2) \ge 8xyz(x+y+z)^3$, which is true because $9(x^2+y^2)(y^2+z^2)(z^2+x^2) \ge 8(x^2+y^2+z^2)(x^2 y^2+y^2 z^2+z^2 x^2)$ and $x^2 y^2+y^2 z^2+z^2 x^2 \ge xyz(x+y+z)$, and we are done. Equality holds iff $x=y=z=3$ and $a=b=c$.

By using Cauchy-Schwarz Inequality, we have \[\sum_{cyc} \frac{\frac{x^2+y^2}{z}}{\frac{1-a^z}{z}}\geq \frac{\left(\frac{x+y}{\sqrt{z}}+\frac{y+z}{\sqrt{x}}+\frac{z+x}{\sqrt{y}}\right)^2}{2\left(1-\frac{a^z}{z}-\frac{b^x}{x}-\frac{c^y}{y}\right)}\]Now, note that \[\frac{a^z}{z}+\frac{b^x}{x}+\frac{c^y}{y}\geq (a^z)^{\frac{1}{z}}(b^x)^{\frac{1}{x}}(c^y)^{\frac{1}{y}}=abc\]by Weighted AM-GM. So, \[\sum_{cyc} \frac{\frac{x^2+y^2}{z}}{\frac{1-a^z}{z}}\geq \frac{\left(\frac{x+y}{\sqrt{z}}+\frac{y+z}{\sqrt{x}}+\frac{z+x}{\sqrt{y}}\right)^2}{2\left(1-\frac{a^z}{z}-\frac{b^x}{x}-\frac{c^y}{y}\right)}\geq \frac{\left(\frac{x+y}{\sqrt{z}}+\frac{y+z}{\sqrt{x}}+\frac{z+x}{\sqrt{y}}\right)^2}{2\left(1-abc\right)}\]and now it is suffices to show that \[\left(\frac{x+y}{\sqrt{z}}+\frac{y+z}{\sqrt{x}}+\frac{z+x}{\sqrt{y}}\right)^2 \geq 12(x+y+z).\]By AM-GM, \[\left(\frac{x+y}{\sqrt{z}}+\frac{y+z}{\sqrt{x}}+\frac{z+x}{\sqrt{y}}\right)^2 \geq 4\left(\sqrt{\frac{xy}{z}}+\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}\right)^2 = 4\left(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}+2(x+y+z)\right)\]and so we need to show that \[\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\geq x+y+z \iff (xy)^2+(yz)^2+(zx)^2\geq xyz(x+y+z)\]which follows from summing up \[x^2y^2+x^2y^2+x^2z^2+x^2z^2\geq 4x^2yz\]\[y^2z^2+y^2z^2+y^2x^2+y^2x^2\geq 4y^2zx\]\[z^2x^2+z^2x^2+z^2y^2+z^2y^2\geq 4z^2yx. \blacksquare\]