$$a^x+b^y+c^z\ge \frac{4abcxyz}{(x+y+z-3)^2} \Longleftrightarrow \frac{1}{xyz}\left(x\left(1-\frac{1}{x}\right)+y\left(1-\frac{1}{y}\right)+z\left(1-\frac{1}{z}\right)\right)^2 \ge \frac{4abc}{a^x+b^y+c^z}$$$$\frac{1}{xyz}\left(x\left(1-\frac{1}{x}\right)+y\left(1-\frac{1}{y}\right)+z\left(1-\frac{1}{z}\right)\right)^2 \stackrel{\text{Weighted AM-GM}}{\ge} \frac{1}{xyz} \left (2\sqrt{ x^{1-\frac{1}{x}}y^{1-\frac{1}{y}}z^{1-\frac{1}{z}} } \right)^2\frac{abc}{abc}=\frac{4abc}{ax{^\frac{1}{x}}by{^\frac{1}{y}}cz{^\frac{1}{z}}} \implies$$$$\frac{1}{xyz}\left(x\left(1-\frac{1}{x}\right)+y\left(1-\frac{1}{y}\right)+z\left(1-\frac{1}{z}\right)\right)^2 \ge \frac{4abc}{ax{^\frac{1}{x}}by{^\frac{1}{y}}cz{^\frac{1}{z}}} \stackrel{\text{Weighted AM-GM}}{\ge} \frac{4abc}{\frac{1}{x}xa^x+\frac{1}{y}yb^y+\frac{1}{z}zc^z}= \frac{4abc}{a^x+b^y+c^z} \ \ \ \blacksquare$$

Another approach (of using weighted A.M-G.M twice): \begin{align*} a^x+b^y+c^z &=xyz\left(\frac{1}{x}\cdot\frac{a^x}{yz}+\frac{1}{y}\cdot\frac{a^y}{zx}+\frac{1}{z}\cdot\frac{a^z}{xy}\right)\\ &\geqslant xyz\left(\frac{a^x}{yz}\right)^{\frac{1}{x}}\left(\frac{a^y}{zx}\right)^{\frac{1}{y}}\left(\frac{a^z}{xy}\right)^{\frac{1}{z}}\\ &=\frac{abcxyz}{(yz)^{\frac{1}{x}}(zx)^{\frac{1}{y}}(xy)^{\frac{1}{z}}}\\ &=\frac{abcxyz}{\left(x^{\frac{1}{2y}}y^{\frac{1}{2z}}z^{\frac{1}{2y}}y^{\frac{1}{2x}}z^{\frac{1}{2y}}x^{\frac{1}{2z}}\right)^2}\\ &\geqslant \frac{abcxyz}{\left(\frac{x}{2y}+\frac{y}{2z}+\frac{z}{2x}+\frac{y}{2x}+\frac{z}{2y}+\frac{x}{2z}\right)^2}\\ &=\frac{4abcxyz}{\left(\frac{x^2y+xy^2+y^2z+yz^2+z^2x+zx^2}{xyz}\right)^2}\\ &=\frac{4abcxyz}{\left(\frac{x^2y+xy^2+y^2z+yz^2+z^2x+zx^2+3xyz}{xyz}-3\right)^2}\\ &=\frac{4abcxyz}{\left(\frac{(x+y+z)(xy+yz+zx)}{xyz}-3\right)^2}\\ &=\frac{4abcxyz}{(x+y+z-3)^2}. \end{align*}

We want $:=$ $$a^x+b^y+c^z\ge \frac{4abcxyz}{(x+y+z-3)^2}.$$$$\iff \sum_{cyc} \frac 1x \cdot \frac {a^x}{yz} \geq \frac {4abc}{(x+y+z-3)^2}$$ Now by Weighted AM-GM we have the estimate $:=$ $$\sum_{cyc} \frac 1x \cdot \frac {a^x}{yz} \geq \left(\frac {a^x}{yz}\right)^{\frac 1x}\cdot \left(\frac {b^y}{zx}\right)^{\frac 1y} \cdot \left(\frac {c^z}{xy}\right)^{\frac 1z}= \frac {abc}{\displaystyle\prod_{cyc} x^{1-\frac 1x}}$$ Hence we will prove the inequality $:=$ $$ \frac {abc}{\displaystyle\prod_{cyc} x^{1-\frac 1x}}\geq \frac{4abc}{(x+y+z-3)^2} \iff (x+y+z-3)^2 \geq 4 \prod_{cyc} x^{1-\frac 1x}$$ Next since $\sum_{cyc} \frac 1x=1$ , hence we can substitute $x=1+ \frac {q+r}{p}$ and so on . We therefore need to prove $:=$ $$\left(\sum_{cyc}\frac {q+r}{p} \right)^2 \geq 4 \prod_{cyc} \left(\frac {p+q+r}{r}\right)^{\frac{q+r}{p+q+r}}$$$$\iff \left(\sum_{cyc}\frac {q+r}{p} \right)\geq 2 \prod_{cyc} \left(\frac {p+q+r}{p}\right)^{\frac{q+r}{2(p+q+r)}}$$ However note that by weighted AM-GM we have $:=$ $$2 \prod_{cyc} \left(\frac {p+q+r}{p}\right)^{\frac{q+r}{2(p+q+r)}}\leq 2 \sum_{cyc} \frac {q+r}{2(p+q+r)} \cdot \frac {p+q+r}{p} = \sum_{cyc} \frac {q+r}{p}$$ This is exactly what we wanted to prove and we are done .

#5 (Algebra) Quite Easy for A3 in my opinion By Weighted AM-GM, \begin{align*} \left(\frac{1}{x}\right)\left(\frac{a^x}{yz}\right)+\left(\frac{1}{y}\right)\left(\frac{b^y}{zx}\right)+\left(\frac{1}{z}\right)\left(\frac{c^z}{xy}\right) &\geq \frac{a}{(yz)^{\frac{1}{x}}}\cdot \frac{b}{(zx)^{\frac{1}{y}}}\cdot \frac{c}{(xy)^{\frac{1}{z}}} \\ \frac{a^x+b^y+c^z}{xyz} &\geq \frac{abc}{x^{\frac{1}{y}+\frac{1}{z}}\cdot y^{\frac{1}{z}+\frac{1}{x}}\cdot z^{\frac{1}{x}+\frac{1}{y}}} \\ a^x+b^y+c^z &\geq \frac{abcxyz}{x^{\frac{1}{y}+\frac{1}{z}}\cdot y^{\frac{1}{z}+\frac{1}{x}}\cdot z^{\frac{1}{x}+\frac{1}{y}}} . \end{align*}Hence, it's suffice to prove that $(x+y+z-3)^2\geq 4 \cdot x^{\frac{1}{y}+\frac{1}{z}}\cdot y^{\frac{1}{z}+\frac{1}{x}}\cdot z^{\frac{1}{x}+\frac{1}{y}}$. By Weighted AM-GM, again, \begin{align*} \left(\frac{\frac{1}{y}+\frac{1}{z}}{2}\right)(x)+\left(\frac{\frac{1}{z}+\frac{1}{x}}{2}\right)(y)+\left(\frac{\frac{1}{x}+\frac{1}{y}}{2}\right)(z) &\geq x^{\frac{\frac{1}{y}+\frac{1}{z}}{2}}\cdot y^{\frac{\frac{1}{z}+\frac{1}{x}}{2}}\cdot z^{\frac{\frac{1}{x}+\frac{1}{y}}{2}} \\ \left(\frac{1}{2}\cdot\left(\frac{x}{y}+\frac{x}{z}+\frac{y}{z}+\frac{y}{x}+\frac{z}{x}+\frac{z}{y}\right)\right)^2&\geq x^{\frac{1}{y}+\frac{1}{z}}\cdot y^{\frac{1}{z}+\frac{1}{x}}\cdot z^{\frac{1}{x}+\frac{1}{y}} \end{align*}Note that the LHS is actually $\left(\frac{1}{2}\cdot\left((x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3\right)\right)^2=\frac{(x+y+z-3)^2}{4}$, as desired.

Note that by wamgm: $$x+y+z-3\geq 2\prod_{\text{cyc}} x^{\frac{1-\frac{1}{x}}{2}}\iff x^{x^{-1}} y^{y^{-1}} z^{z^{-1}}\geq \frac{4xyz}{x+y+z-3} $$So by wamgm: $$\sum_{\text{cyc}} \frac{xa^x}{x}\geq abc x^{x^{-1}} y^{y^{-1}} z^{z^{-1}} \geq \frac{4xyzabc}{x+y+z-3}$$

We apply weighted AM-GM on the first expression by taking weights as $\frac1x,\frac1y,\frac1z$. \[a^x+b^y+c^z \geq \left(\frac{a^x}{\frac1x}\right)^{\frac1x}\left(\frac{b^y}{\frac1y}\right)^{\frac1y}\left(\frac{c^z}{\frac1z}\right)^{\frac1z}=abcx^{\frac1x}y^{\frac1y}z^{\frac1z}\]We now show that \[x^{\frac1x}y^{\frac1y}z^{\frac1z}\geq\frac{4xyz}{(x+y+z-3)^2}\]\[\Longleftrightarrow (x+y+z-3)^2\geq4x^{\frac1y+\frac1z}y^{\frac1z+\frac1x}z^{\frac1x+\frac1y}\]However, note that by weighted AM-GM, we do have \[\frac{x\left(\frac1y+\frac1z\right)+y\left(\frac1z+\frac1x\right)+z\left(\frac1x+\frac1y\right)}2=\frac{x\left(\frac1y+\frac1z\right)+y\left(\frac1z+\frac1x\right)+z\left(\frac1x+\frac1y\right)}{2\left(\frac1x+\frac1y+\frac1z\right)}\geq\left(x^{\frac1y+\frac1z}y^{\frac1z+\frac1x}z^{\frac1x+\frac1y}\right)^{\frac1{2\left(\frac1x+\frac1y+\frac1z\right)}}\]We may also write \[x\left(\frac1y+\frac1z\right)+y\left(\frac1z+\frac1x\right)+z\left(\frac1x+\frac1y\right)=x\left(1-\frac1x\right)+y\left(1-\frac1y\right)+z\left(1-\frac1z\right)=x+y+z-3\]Substituting this in the previous equation, we get the desired result. $\blacksquare$