A nonnegative function $f(a,b)\geq 0$ Let $c$ be a fixed positive real  umber such that $c\geq 1.$ If $a\geq c^m, b\geq c, $ where $m$ is any positive integer, then is
$$f(a,b)=c(ab+1)(a-c^m)(b-c)+c(ab+1)(a-c^m)(b+1)+c^m(b-c)(ab+1)(a+1)-(ab-c^{m+1})(a+1)(b+1)\geq 0?$$
We can check the validity for the two  cases $a=c^m,$ or   $b=c$ easily. In fact the equality in the above claim  occurs at  $(a, b)=(c^m,c).$ Kindly suggest me the way  to establish $f(a,b)\geq 0$ if it is true.

Yes, it's true if all my computation is right.
First, the last term can be divided into two part.
$$(ab-c^{m+1})(a+1)(b+1)=(ab-c^{m+1})(ab+1)+(ab-c^{m+1})(a+b) $$
Use the first term to deal the first part and see what remains
$$ c(ab+1)(a-c^m)(b-c)-(ab-c^{m+1})(ab+1)\geq (ab+1)(a-c^m)(b-c)-(ab-c^{m+1})(ab+1)\\
=-(ab+1)(c(a-c^m)+c^m(b-c))$$
Now divide the second and third term in $f$
$$c(ab+1)(a-c^m)(b+1)=cb(ab+1)(a-c^m)+c(ab+1)(a-c^m)$$
$$c^m(b-c)(ab+1)(a+1)=c^ma(b-c)(ab+1)+c^m(b-c)(ab+1)$$
Their second part can fill the difference the first remains and now it's sufficient to prove
$$cb(ab+1)(a-c^m)+c^ma(b-c)(ab+1)-(ab-c^{m+1})(a+b)\geq0 $$
Note the fact $a\geq c^m>1, \; b\geq c \geq 1$, $(ab+1)\geq(a+b)$, so it's sufficient to prove
$$ cb(a-c^m)+c^ma(b-c)-(ab-c^{m+1})\geq 0 $$
Notice
$$cb(a-c^m)+c^ma(b-c)-(ab-c^{m+1})=(c-1)(a-c^m)(b-1)+c^m(a-1)(b-1)+(c-1)a\geq 0$$
And now we know $f$ is always non-negative.
