Equivalent to $(y-x)(ax^2 + ay^2 - x^2 - xy) \geq 0$. For $x=1$ and $y>1$ we would like $ay^2 - y + a - 1 \geq 0$ to hold for all $y>1$ - but if it did, then taking the limit as $y\to 1$ would imply $2(a-1) \geq 0$, i.e. $a\geq 1$, contradiction. Hence no such $a<1$ exists.

I claim there is no such $a$. Suppose the contrary and write $(1-a)x^3 + (1-a)xy^2 + ax^2y+ay^3\ge 2xy^2$, namely, $(1-a)x^3 + ax^2y+ay^3 \ge (1+a)xy^2$. This is equivalent to \[ x(x-y)(x+y)\ge a(x^2+y^2)(x-y). \]Suppose this inequality is valid for all $y>x$. Then, $(1-a)x^2+xy<ay^2$ for all $0<x<y$, which implies $x<ay$. Now, take $x$ such that $ay<x<y$, such $x$ must exist as $a<1$, a clear contradiction.