Show that $4(x^3 + y^3 + z^3 + 3) \geq 3(x+1)(y+1)(z+1)$ | Inequalities

Question

I have the following problem involving inequalities.

Let $x, y, z > 0$ be real numbers. Show that the following is true: $$4(x^3 + y^3 + z^3 + 3) \geq 3(x+1)(y+1)(z+1)$$ For which values of $x, y, z$ does equality hold?

At first I simplified as much as possible: $$4x^3+4y^3+4z^3+12 \geq 3xyz + 3xy + 3xz + 3yz + 3x + 3y + 3z + 3$$

$$4x^3+4y^3+4z^3+9 \geq 3xyz + 3xy + 3xz + 3yz + 3x + 3y + 3z$$

But now I'm stuck. I know that the solution has something to do with geometric and arithmetic means, and that $\frac{a+b}{2}\geq\sqrt{ab}$. I just can't seem to see how to transform the inequality into something like that.

How would you solve such a question? I would greatly appreciate help!

Answer 1

By Holder and AM-GM we obtain:$$4(x^3+y^3+z^3+3)=4(x^3+1)+4(y^3+1)+4(z^3+1)\geq$$ $$\geq(x+1)^3+(y+1)^3+(z+1)^3\geq3(x+1)(y+1)(z+1).$$

Answer 2

Another way.

Let $x+y+z=3u$.

Thus, $$4(x^3+y^3+z^3+3)-3\prod_{cyc}(x+1)=$$$$=\sum_{cyc}(4x^3-xyz)-3(xy+xz+yz)-3(x+y+z)+9\geq$$ $$\frac{1}{3}(x+y+z)^3-(x+y+z)^2-3(x+y+z)+9=$$ $$=9(u^3-u^2-u+1)=9(u-1)^2(u+1)\geq0.$$

Answer 3

Another way.

We need to prove that: $$4(x^3+y^3+z^3)+9\geq3xyz+3(xy+xz+yz)+3(x+y+z)$$ and since by AM-GM $$x^3+y^3+z^3\geq3xyz$$ and $$\sum_{cyc}3xy\leq\sum_{cyc}(x^3+y^3+1)=2(x^3+y^3+z^3)+3,$$ it's enough to prove that: $$x^3+y^3+z^3+6\geq3(x+y+z),$$ which is true by AM-GM: $$\sum_{cyc}(x^3+2)\geq\sum_{cyc}3\sqrt[3]{x^3\cdot1^2}=3(x+y+z).$$

Answer 4

Using Am_Gm inequality we see that

$$(x+1)(y+1)(z+1)\leq {(x+1)^3+(y+1)^3 +(z+1)^3\over 3}$$

so it is enough to check if $$(x+1)^3+(y+1)^3 +(z+1)^3 \leq 4(x^3 + y^3 + z^3 + 1+1+1)$$ is true (by means of elementary inequalities). Actualy it is again enough to check if $$(x+1)^3\leq 4x^3+4 = 4(x+1)(x^2-x+1)$$ is true, i.e. if $$x^2+2x+1\leq 4x^2-4x+4$$ is true.