Proving $6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$, where $x+y+z=0$

Question

Question :

Let $x,y,z$ be reals satisfying $x+y+z=0$. Prove the following inequality:$$6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$$

My Attempts : It’s obvious that $x,y,z$ are either one negative and two positive numbers or one positive and two negative numbers.

In addition, putting $(x,y,z)$ and $(-x,-y,-z)$ into the inequality has the same outcome.

Therefore, without loss of generality, I suppose $x,y \geq 0$ and $z\leq0$.

$x+y=-z\\ \Longrightarrow (x+y)^2=z^2 \\ \Longrightarrow (x^2+y^2+z^2)^3=8(x^2+xy+y^2)^3$

is what I have got so far, and from here I can’t continue.

Am I on the right direction? Any suggestions or hints will be much appreciated.

Answer 1

Now, let $x^2+y^2=2uxy.$

Thus, since $xy\geq0$ and for $xy=0$ our inequality is true, we can assume that $xy>0$, which gives $u\geq1$ and we need to prove that: $$8(x^2+xy+y^2)^3\geq6(-3xy(x+y))^2$$ or $$2(2u+1)^3\geq27(u+1)$$ or $$16u^3+24u^2-15u-25\geq0,$$ which is obvious for $u\geq1$.

Answer 2

$$z=-(x+y)$$

$$\implies 6 [x^3 + y^3 - (x+y)^3]^2 - [x^2 + y^2 +(x+y)^2]^3$$ $$=-8 x^6 - 24 x^5 y + 6 x^4 y^2 + 52 x^3 y^3 + 6 x^2 y^4 - 24 x y^5 - 8 y^6$$ $$=-2\,(x-y)^2 \, (2 x^2 + 5 x y + 2 y^2)^2\leq 0$$

Answer 3

The difference $$ (x^2+y^2+z^2)^3-6(x^3+y^3+z^3)^2 $$ for $x+y+z=0$ can be written $$ 2 (x-y)^2 (y-z)^2 (z-x)^2 $$

Answer 4

Scaling, one may assume that $z = -1$, so that $x + y = 1$. Then the desired inequality is $$6(x^3 + y^3 - 1)^2 \leq (x^2 + y^2 + 1)^3$$ Since $x^3 + y^3 = (x + y)^3 - 3xy(x + y) = 1 - 3xy$ and $x^2 + y^2 + 1 = (x + y)^2 - 2xy + 1 = 2 - 2xy$, the desired inequality is $$54(xy)^2 \leq (2 - 2xy)^3$$ Thus it makes sense to look at $xy = x(1 - x) = {1 \over 4} - (x+{1 \over 2})^2$, whose range is $(-\infty, {1 \over 4}]$. Letting $r = xy$, we need that for $r \leq {1 \over 4}$ we have $$54r^2 \leq (2 - 2r)^3$$ However, the polynomial $54r^2 - (2 - 2r)^3$ can be directly factorized into $2(r + 2)^2(4r - 1)$, which is nonpositive in the domain $(-\infty, {1 \over 4}]$. Hence the inequality holds.

Answer 5

Thanks for all the answers you guys provided, and after spending a large amount of time wandering in this question and understanding these answers, I came up with my solution which I think is valid.

By AM-GM:

$x^2+xy+y^2\\=\frac{x(x+y)}{2}+\frac{y(x+y)}{2}+\frac{x^2+y^2}{2}\\ \geq3\cdot \sqrt[3]{\frac{xy(x+y)^2}{4}\cdot\frac{x^2+y^2}{2}}\\\geq 3\cdot \sqrt[3]{\frac{x^2y^2z^2}{4}}$

Therefore $(x^2+y^2+z^2)^3\geq8\cdot(3\cdot \sqrt[3]{\frac{x^2y^2z^2}{4}} )^3=54x^2y^2z^2$

Now, since $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$ and $x+y+z=0$,

we have $6(x^3+y^3+z^3)^2=6(3xyz)^2=54x^2y^2z^2$

Hence the inequality is proven.