Prove $4(x^2+1)^3 \leq 5(x^3+2)^2$ I need to prove that $4(x^2+1)^3 \leq 5(x^3+2)^2$ ,  for all $x\geq 0$, in the context of an introductory analysis course. I thought I could somehow use convex functions to prove this, but I simply cannot get it to work. Hints anyone?
Hint :
$$ 5(x^3+2)^2 - 4(x^2+1)^3 = (x-2)^2(x^4+4x^3+4x+4)$$
we must prove :
$$4x^6+12x^4+12x^2+4 \leq 5x^6+20x^3+20$$
which rewrites to :
$$x^6-12x^4+20x^3-12x^2+16 \ge0$$
or :
$$(x-2)(x^5+2x^4-8x^3+4x^2-4x-8) \geq 0$$
or :
$$(x-2)^2(x^4+4x^3+4x+4)\geq 0$$
which is clearly true, with equality at $x=2$