Consider the polynomial
$$ P(x,y)=15 \times 10^{15} (x^4+y^4)+2 \times 9999^2 \times 10^{12} xy(x+y-3)+20001 \times 9999^4 $$
One can show that $P(x,y)\geq 0$ when $x,y\geq 0$ (proof is below). Since $P$ has only one negative coefficient, one would expect that there is a simple proof of this nonnegativity property of $P$.
However, so far I only found the ugly proof at the end of this question which uses integers with more than $20$ digits.
My question: Is there a simpler method ?
My thoughts: The middle part of $P$ resembles the Motzkin polynomial ; if we put $m(x,y)=xy(x+y-3)+1$ we have $m(x,y)\geq 0$ by AM-GM. This suggest rewriting $P$ as $P(x,y)=15 \times 10^{15}(x^4+y^4)+2 \times 9999^2 \times 10^{12}m(x,y)-3\times 9999^2 \times 10^8$, but unfortunately we still have a negative coefficient.
Here is my painstaking proof. Let $r=1-\frac{1}{10^4}$ (note that $P(r,r)=0$) and $l=r+\frac{3332}{6669}(r-x)$.
When $x\leq 4$, we have $P(x,y)\geq 0$ because
$$ \begin{array}{rl} P(x,y)= & \bigg(c_1+c_2(4-x)+c_3(4-x)^2+\big(c_4+c_5(4-x)\big)y\bigg)(x-r)^2 \\ & + (c_6+c_7l^2)(y-l)^2+c_8(y^2-l^2)^2 \end{array} $$
where
$$ \begin{array}{c} c_1=\frac{1632689280798968825215\times 10^7}{32043471291},c_2=\frac{ 30294845044444411219\times 10^{12}}{608825954529}, \\ c_3=\frac{2680495655043655\times 10^{15}}{219786169584969},c_4=\frac{897083745816842\times 10^{12}}{5203640637},c_5=\frac{73985220736\times 10^{16}}{98869172103},\\ c_6= 199960002\times 10^{12}, c_7=3\times 10^{16}, c_8=15\times 10^{15} \end{array} $$
So we may assume WLOG that $x\gt 4$. By symmetry, we may also assume $y\gt 4$. Then $x+y-3\geq 5$, the middle part of $P$ is positive and we are done.
If $x + y \ge 3$, clearly $P(x, y) \ge 0$.
If $x + y < 3$, using $x^4 + y^4 \ge (x + y)^4/8$ (power mean inequality) and $xy \le \frac{(x+y)^2}{4}$, it suffices to prove that $$15 \times 10^{15} \cdot \frac{(x+y)^4}{8} + 2 \times 9999^2 \times 10^{12} \cdot \frac{(x+y)^2}{4} \cdot (x + y - 3) + 20001 \times 9999^4 \ge 0.$$
It suffices to prove that, for all $u\ge 0$, $$15 \times 10^{15} \cdot \frac{u^4}{8} + 2 \times 9999^2 \times 10^{12} \cdot \frac{u^2}{4} \cdot (u - 3) + 20001 \times 9999^4 \ge 0$$ or (multiplying both sides by $16/9999^4$) $$3\left(\frac{10^4u}{9999}\right)^4 + 8\times 9999\times \left(\frac{10^4u}{9999}\right)^3 - 24 \times 10^4 \left(\frac{10^4u}{9999}\right)^2 + 16\times 20001 \ge 0.$$
It suffices to prove that, for all $v \ge 0$, $$3v^4 + 8\times 9999 v^3 - 24 \times 10^4v^2 + 16 \times 20001 \ge 0$$ or (letting $a = 9999$) $$3v^4 + 8av^3 - 24(1 + a)v^2 + 16(2a + 3) \ge 0$$ or $$(8v^3 - 24v^2 + 32)a + 3v^4 - 24v^2 + 48 \ge 0$$ or $$8(v+1)(v-2)^2 a + 3(v^2 - 4)^2 \ge 0$$ which is clearly true.
We are done.