$x^6+5x^5y+10x^4y^2+kx^3y^3+10x^2y^4+5xy^5+y^6\ge 0$

Find the absolute value of the smallest possible $k$ such that the inequality above is true for all non-negative reals $x$ and $y$ .

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Note:
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You may use the algebraic identities below.

- $(x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$
- $(x+y)^6=x^6+6 x^5 y+15 x^4 y^2+20 x^3 y^3+15 x^2 y^4+6 x y^5+y^6$

The answer is 32.

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Sorry to say, the entire Details and Assumptions, along with the title, some of the tags, and the problem itself, were just red herrings, trying to persuade you that Pascal's Triangle and the Binomial Theorem had something to do with it. The solution (at least the intended solution) has nothing to do with them, at all.

First, move the $kx^3y^3$ term to the right side of the equation. We can just let the resulting $-k$ coefficient be $k$ since we'll be taking the absolute value of it anyways at the end. $x^6+5x^5y+10x^4y^2+10x^2y^4+5xy^5+y^6\ge kx^3y^3$

Now, note that $(3,3)\prec (6,0),(5,1),\text{ and }(4,2)$ ; thus, by Muirhead's Inequality, $x^6+y^6\ge 2x^3y^3$ $x^5y+xy^5\ge 2x^3y^3$ $x^4y^2+x^2y^4\ge 2x^3y^3$

This means $(x^6+y^6)+5(x^5y+xy^5)+10(x^4y^2+x^2y^4)\ge 2x^3y^3 + 5(2x^3y^3)+10(2x^3y^3)=32x^3y^3$

Thus, our answer is $\boxed{32}$ .