A Relative of IMO Classic $a^2+b^2=c^2(1+ab)$

Question

If $a,b,c \in \mathbb{N}$ such that $$a^2+b^2=c^2(1+ab)$$ Then prove that $$a \geq c \: \text{and} \:b \geq c$$

My try:

{Case I:Let $a=b$} Then we have $$2a^2=c^2+c^2a^2 \Rightarrow a^2(2-c^2)=c^2 \Rightarrow c^2<2 \Rightarrow c=1$$ $$c=1 \Rightarrow a=b=1 \Rightarrow a=c,b=c$$

{Case II:Let $a \ne b$}:

{SubCase I:If $c=1$} Then we have $$a^2+b^2=1+ab \Rightarrow (a-b)^2=1-ab \Rightarrow ab<1 \Rightarrow a,b \notin \mathbb{N}$$ So this subcase is Rejected.

{SubCase II: $c \geq 2$ and also let $a>b$} We get $$(a-b)^2=c^2+(c^2-2)ab \Rightarrow (a-b)^2-c^2=(c^2-2)ab$$ So since $c^2-2>0$ $$(a-b)^2-c^2 >0 \Rightarrow a-b>c \Rightarrow a>b+c>c \Rightarrow a>c$$

{Subcase III:$c \geq 2$ and also $b>a$} Following the same lines as above, we get $b>c$.

Now how can I prove both $a>c,b>c$ holds simultaneously?

Does this answer your question? $x^2+y^2=z^2(1+xy)$ prove $z=\min {x;y;z}$ (with $x,y,z \in \mathbb{Z^+}$) - found using an Approach0 search. The duplicate's answer uses an apparently different approach, so it doesn't contain how to finish your method. ... — John Omielan, Nov 10 '22 at 02:33
(cont.) However, zyx's question comment states that "It contains the non-Diophantine problem of seeing that a proof by pure inequalities cannot work." I'm unclear if this is referring to a proposed duplicate involving the related, and classic, problem of showing that if $\frac{a^2+b^2}{ab+1}$ is an integer, it must also be a perfect square. If it's referring to your specific problem, then this suggests there might not be any way, at least any easy one, to finish your approach. — John Omielan, Nov 10 '22 at 02:39

A Relative of IMO Classic $a^2+b^2=c^2(1+ab)$

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