Prove for $abc$-triples that $c\leq rad(abc)^2$

Question

Prove for $abc$-triples that $c\leq \text{rad}(abc)^2$. $\text{rad}(abc)$ and $\text{rad}(x)$ means here the product of all prime factors of $x$.

(edit 2)

The above holds for:

$16+5=21 \leq (2*5*3*7)^2$

$3^2+4^2=5^2 \leq (3*2*5)^2$

$1+99=100 \leq (2*11)^2$

I am looking for a general condition which would at least satisfy the above examples.

Answer 1

Currently it is unkwown whether or not there exists an $(a,b,c)$-triple, i.e., coprime numbers $a,b,c$ with $a+b=c$ and $c\ge a,b$, such that $$ c>rad(abc)^2. $$ So the formulation "prove that it holds" is quite unrealistic. For references on these claims see, say, here. This is closely related to the so-called "explicit" $abc$-conjecture.