Wow, I really like this problem. Congrats Shantanu!

Solved with AlperenInan. Here goes a bashy one. Let $a=dyzk, \ b=dxzl, \ c=dxym$ such that $(b,c)=dx,(a,c)=dy,(a,b)=dz$ and $(k,l)=(l,m)=(m,k)=1$. Note that this requires $(x,y)=(y,z)=(z,x)=1$ since $(x,y,z)=d$. \[d^3x^2y^2z^2klm|d^4z^4y^3xk^3l+d^4x^4z^3l^3ym+d^4y^4x^3m^3zk\iff xyzklm|d(z^3k^3y^2l+x^3l^3z^2m+y^3m^3x^2k)\]We see that $x|dz^3k^3y^2l$ which is equavilent to $x|d$. Since $(x,y)=(y,z)=(z,x)=1$ similarily $xyz|d$. Let's write $d=xyzp$. \[klm|p(z^3k^3y^2l+x^3l^3z^2m+y^3m^3x^2k)\]$k|px^3l^3z^2m$ thus, $k|pz^2$. WLOG $a=\min\{a,b,c\}$ and $a\neq b\neq c\neq a$. Also by the given condition, \[dyzk+\sqrt{\frac{dyzk}{2}}>dxzl\iff yk+\sqrt{\frac{yk}{2dz}}>xl\geq yk+1\implies yk\geq 2dz=2xyz^2p\implies k\geq 2xz^2p\]Also note that $k\leq pz^2$. However $pz^2\geq k\geq 2x(pz^2)>pz^2$ is impossible. Thus, the assumption $a\neq b\neq c\neq a$ does not hold. So suppose that $b=c$. We have $a|b^2$. Now write $a=dx,b=dy$ such that $(a,b)=d$. Since $dx|d^2y^2$ and $(x,y)=1$, we see that $x|d$. Let $d=xq$. \[x^2q+x\sqrt{\frac{q}{2}}>xyq\implies x+\sqrt{\frac{1}{2xq}}>y\implies x\geq y\]\[xyq+\sqrt{\frac{xyq}{2}}>x^2q\implies y+\sqrt{\frac{y}{2xq}}>x\implies y\geq x\]As desired.$\blacksquare$

Nothing!

Solved with Lleeya. I think starchan's solution is the intended one. XD. We will solve the strengthened version where $a,b,c \in \left[ M, M + 3 M^{\frac{3}{4}} \right]$, as restated below. Stronger Form wrote: Let $M$ be a positive integer, and let $a,b,c$ be integers in the interval $\left[M,M+3 M^{\frac{3}{4}} \right]$ such that $a^3b+b^3c+c^3a$ is divisible by $abc$. Prove that $a=b=c$. Originally proposed by Shantanu Nene First, for any prime $p \mid a$, we have that $p \mid b^3c$ which implies that $p \mid b$ or $p \mid c$. This means that for each prime factor of $abc$, at most one of $a,b,c$ are not divisible by the prime factor. Pick any prime factor $p$ of $\gcd (a,b)$. From $ab \mid abc \mid a^3b + b^3c + c^3a$, we have that $\nu_p(ab) \le \nu_p(c(b^3 + c^2a))$. However, if $p \nmid c$, we get that $\nu_p(a) + \nu_p(b) \le \nu_p(b^3 + c^2a)$. If $3\nu_p(b) = \nu_p(b^3) > \nu_p(c^2a) = \nu_p(a)$, we have that $\nu_p(b^3 + c^2a) = \nu_p(a)$, a clear contradiction. Hence, we must have that $3\nu_p(b) \le \nu_p(a)$. Also, when $3\nu_p(b) = \nu_p(b^3) < \nu_p(a)$, we have that $\nu_p(b^3 + c^2a) = \nu_p(b^3) > 4\nu_p(b)$, a contradiction. Hence, we must have $\nu_p(a) = 3\nu_p(b)$, and it goes same with other prime factors of $\gcd(b,c)$ and $\gcd(c,a)$ that does not divide $\gcd(a,b,c)$. Therefore, there exists pairwise coprime integers $x,y,z$ so that $(a,b,c) = (dxy^3, dyz^3, dzx^3)$ when $d = \gcd(a,b,c)$. Now, wlog assume that $z = \max \{ x, y, z \}$. The given condition becomes $xyz \mid d(x^7 + y^7 + z^7)$, but we will not use this from now on because controlling $d$ will force this to satisfy, but it is harder to bound $d$. Here, I suddenly changed the cyclic form, but it does not matter. Sorry for the inconvenience. It became $(x^3y, y^3z, z^3x)$. Since $z^3x \ge x^3y$ from the assumption, we have that $z^3x \in \left[ x^3y , x^3y + 3 (x^3y)^{\frac{3}{4}} \right]$. Why? This holds since dividing $d$ gives a stricter bound We have three cases where $x > y$, $x < y$, and $x=y=1$. If $x > y$, we have that $z^3x \le x^3y + 3(x^3y)^{\frac{3}{4}} < x^4 + 3x^3$. This implies that $z^3 < x^3 + 3x^2 < (x+1)^3$. Hence, $z = x$, but since $z$ and $x$ are coprime, we have $x=z=1>y$, a contradiction. If $y > x$, we have that $z^3x \le x^3y + 3x^{\frac{9}{4}}y^{\frac{3}{4}}$. This implies that $z^3 \le x^2y + 3x^{\frac{5}{4}}y^{\frac{3}{4}}$. However, since $z > y$ or $z=y=1 > x$, we have that $y \le z-1$. Plugging this bound gives $z^3 < (z-1)^3 + 3(z-1)^2 < z^3$, a contradiction. Therefore, both cases have no solution, we can conclude that $x=y=1$. This implies that $z^3 \in [1, 4]$, or, $z = 1$. This case gives $a=b=c$, as desired.