[Polynomials] Showing the Motzkin Polynomial is Non-Negative using the AM-GM Inequality

Let's take a look at the Motzkin polynomial P(x,y)=x4y2+x2y4+13x2y2.P(x,y) = x^4y^2+x^2y^4 +1-3x^2y^2. How can we show that this polynomial satisfies P(x,y)0P(x,y) \geq 0 on R2\mathbb{R^2}?

We can look to the arithmetic mean-geometric mean (AM-GM) inequality to show this. The AM-GM inequality states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list. Applying this inequality to the list of numbers x2y4,x4y2,1x^2y^4, x^4y^2, 1, we can compute the arithmetic mean x2y4+x4y2+13,\frac{x^2y^4 + x^4y^2 +1}{3}, and the geometric mean (x2y4)(x4y2)13=x6y63=x2y2.\sqrt[3]{(x^2y^4)(x^4y^2)\cdot 1} = \sqrt[3]{x^6y^6} = x^2y^2.

Hence the inequality gives x2y4+x4y2+13x2y2\frac{x^2y^4 + x^4y^2 +1}{3} \geq x^2y^2 Multiplying both sides by 3 and rearranging, x2y4+x4y2+13x2y20 x^2y^4 + x^4y^2 +1 - 3x^2y^2 \geq 0 as required.

We see that the AM-GM inequality provides a simple yet effective way to show whether polynomials over a real field are non-negative (or non-positive). You can also try showing that the polynomial Q(x,y)=2x4+5y4x2y2+2x3y Q(x,y) = 2x^4+5y^4-x^2y^2+2x^3y over a real field, is non-negative as an exercise.

Note by Bright Glow
3 years, 2 months ago

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