Conjecture: If $A, B, C$ are positive integers with $\gcd(A, B)=1$, $\gcd(B, C)=1$, $\gcd(C, A)=1$. If $A+B=C$ then $min(A,B) \le rad(ABC)$

If the conjecture is valid, we can use the conjecture to prove the Fermat last theorem as follows:

Proof the Fermat last theorem:

We consider the Fermat equation:

$x^n+y^n=z^n$ with $\gcd(x, y)=1$, $\gcd(y, z)=1$, $\gcd(z, x)=1$

There is no loss of generality in assuming that $x \le y <z$. By the conjecture we get $x^n \le rad(x^n.y^n.z^n)=rad(x.y.z)\le zyz $.

So $x^n+y^n \le xyz+y^n<z^3+(z-1)^n$

But easily we can prove that $z^3+(z-1)^n < z^n$ with $n > 3$ and $z>1$. So now we only need to prove the Fermat last theorem with $n=3$.

My question: Is the conjecture above new and correct?

$\DeclareMathOperator\rad{rad}$Conjecture: If $A, B, C$ are positive integers with $\gcd(A, B)=1$, $\gcd(B, C)=1$, and $\gcd(C, A)=1$, and if $A+B=C$, then $\min(A,B) \le \rad(ABC)$.

If the conjecture is valid, then we can use the conjecture to prove the Fermat last theorem as follows:

Proof of the Fermat last theorem:

We consider the Fermat equation:

$x^n+y^n=z^n$ with $\gcd(x, y)=1$, $\gcd(y, z)=1$, and $\gcd(z, x)=1$.

There is no loss of generality in assuming that $x \le y <z$. By the conjecture, we get $x^n \le \rad(x^n y^n z^n)=\rad(x y z)\le x y z $.

So $x^n+y^n \le xyz+y^n<z^3+(z-1)^n$.

But we can easily prove that $z^3+(z-1)^n < z^n$ whenever $n > 3$ and $z>1$. So now we only need to prove the Fermat last theorem with $n=3$.

My question: Is the conjecture above new and correct?