Let $D,E,F,G$ be points such that \[\triangle ABD\sim\triangle ADE\sim\triangle AEF\sim\triangle AFG\sim\triangle AGC \]Then $K$ lies on $AF$. Then \[\angle BDE=\angle DEF=\angle EFG=\angle FGC=180-\angle BAD<180 \] so $D,E,F,G$ are all outside the triangle. Note that $AB^2AC^3=AF^5>AK^5$, so we are done.

It's worth noting that this problem was proposed by the one and only $\textit{Nikolai Beluhov}$. The official solution is the one presented in #2. Here's something I consider much more intuitive: $\textbf{Lemma:}$ Let $DL$ be the angle bisector in $\triangle DEF$. Then $DE\cdot DF>DL^2$ $\textit{Proof:}$ Let $DE=f$, $DF=e$, $EL=m$, $LF=n$, then by Stewart's theorem: \[DL^2=\frac{nf^2+me^2}{m+n}-mn=ef-mn<ef=DE\cdot DF\]where we used the Angle Bisector theorem to get $\frac{nf^2+me^2}{m+n}=ef$. Now, let's construct points $X,Y,Z\in BC$, such that $\angle BAX=\angle XAY=\angle YAK=\angle KAZ=\angle ZAC$. This is possible because $2\angle BAK=3\angle KAC$. Denote $AB=c$, $AC=b$, $AX=x$, $AY=y$, $AZ=z$, $AK=k$. Then, using the Lemma, we have that: \[k^2<bx=\frac{b}{x^{\frac{1}{3}}}\cdot x^{\frac{4}{3}}<\frac{b}{x^{\frac{1}{3}}}\cdot (c^{\frac{2}{3}}y^{\frac{2}{3}})<\frac{bc^{\frac{2}{3}}}{x^{\frac{1}{3}}}\cdot (x^{\frac{1}{3}}k^{\frac{1}{3}})=bc^{\frac{2}{3}}\Longrightarrow k^5<b^3c^2\Longrightarrow AK^5<AB^2AC^3\]

Here is a solution using a very neat lemma Let $x = \angle BAK$ and $y=\angle KAC$, from here we have that $y=\frac{2}{5}\alpha$ and $x=\frac{3}{5}\alpha$. Using the law of sines andrearranging the ineqluatiy we need to show that: $$\sin^5\left( \frac{2}{5}\alpha + \gamma \right) > \sin^3 \gamma \sin^2 (\alpha + \gamma)$$which brings us to the following lemma. Lemma: Let $x$ and $y$ be arbitrary real numbers and $\alpha$ and $\beta$ some angles in the interval $(0,\pi)$, then we have that: $$\sin^x \alpha \sin^y \beta < \sin^{x+y}\frac{x\alpha+y\beta}{x+y}$$Proof: Put both sides under the natual log, and let $f(x)=\ln\sin x$, then we must show that: $$\frac{x}{x+y}f(\alpha)+\frac{y}{x+y}f(\beta) < f \left( \frac{x\alpha+y\beta}{x+y} \right)$$which would imply that $f$ is concave, but notice that $f''(x)=-\frac{1}{\sin^2 x}$, which is always strictly less than $0$, thus proving the upper inequality. $\blacksquare$ Now applying our lemma we get the desired result, i.e. we have that: $$\sin^5\left( \frac{2}{5}\alpha + \gamma \right) > \sin^3 \gamma \sin^2 (\alpha + \gamma)$$which proves $AB^2AC^3 > AK^5$.

Rename $K$ to $D$. Suppose $\angle BAD=3x,\angle CAD=2x$, so $\angle BAC=5x$. Let $\overline{AE}$ bisect $\angle CAD$, so $\angle CAE=\angle EAD=x$, thus $\angle BAE=4x$. Now let $\overline{AF}$ bisect $\angle BAE$, so $\angle BAF=2x$ and thus $\angle FAD=x$. Thus $\overline{AD}$ is the bisector of $\angle FAE$ as well since $\angle FAD=\angle EAD=x$. Using the $\sqrt{ab-xy}$ "AIME Geo formula" for the length of a bisector, we find that the length of the bisector of a triangle is strictly less than the square root of the product of the sides surrounding it. Let $AB=x,AC=y,AD=z$. Then $$AE<\sqrt{yz}, AF<\sqrt{x\sqrt{yz}} \implies z=AD<\sqrt{AE\cdot AF}<\sqrt{\sqrt{yz}\cdot \sqrt{x\sqrt{yz}}}.$$Thus, $$z^2<\sqrt{yz}\cdot \sqrt{x\sqrt{yz}} \implies z^4<xyz\sqrt{yz} \implies z^6<x^2y^3z \implies z^5<x^2y^3,$$as desired.