Prove that $3 a^4+ 2 a^3 b - 3 a^2 b^2 - 2 a b^3 + 3 b^4\geq 0$.

Question

Let $a,b\geq 0$. How can I prove $$3 a^4+ 2 a^3 b - 3 a^2 b^2 - 2 a b^3 + 3 b^4\geq 0$$ ?

I try using Schur and Muirhead but they didn't work here because Schur is for three variables and Muirhead works for symmetric sums...

Answer 1

By AM-GM we have $\frac32 (a^4+b^4)\geq 3a^2b^2$ so it is enough to prove that $$\frac32(a^4+b^4)+2a^3b\geq 2ab^3.$$

Your inequality is not very sharp so we can just drop some terms: $$\frac32(a^4+b^4)+2a^3b\geq\frac32(a^4+b^4)\geq\frac12(a^4+b^4+b^4+b^4)\overset{\text{AM-GM}}\geq\frac42\sqrt[4]{a^4b^{12}}=2ab^3.$$

Answer 2

If either $a = 0$ or $b = 0$, the inequality is trivially true. So assume otherwise that $a > 0, b > 0$. Divide both sides by $b^4$ we have an equivalent inequality: $3x^4 +2x^3 - 3x^2 - 2x + 3 > 0$ with $x = \dfrac{a}{b} > 0$ Divide both sides of this inequality by $x^2$ we have: $3x^2 + 2x - 3 - \dfrac{2}{x} + \dfrac{3}{x^2} > 0$. Now put $t = x - \dfrac{1}{x}$, then we prove: $3t^2+2t+3 > 0$ and this is clearly true for all real number $t$. Thus the above inequality is true.

Answer 3

Let $a^2-b^2=2kab$.

Thus, we need to prove that: $$a^2b^2(3(4k^2+2)+2\cdot2k-3)\geq0$$ or $$12k^2+4k+3\geq0,$$ which is true because $2^2-12\cdot3<0.$