Prove that $(x + y + z)^3 + 9xyz \ge 4(x + y + z)(xy + yz + zx)$ where $x, y, z \ge 0$.

Question

Prove that $(x + y + z)^3 + 9xyz \ge 4(x + y + z)(xy + yz + zx)$ where $x, y, z \ge 0$.

This has become the norm now... This problem is adapted from a recent competition.

We have that $6(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) \ge 4(x + y + z)(xy + yz + zx)$

Furthermore, $$(x + y + z)^3 - 6(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) + 9xyz$$

$$ = x^3 + y^3 + z^3 - 3(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) + 15xyz$$

In addition, $x^3 + y^3 + z^3 + 3xyz \ge x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2$.

So now we need to prove that $6xyz \ge x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2$, which I don't even know if it is true or not.

Answer 1

We need to prove that $$\sum_{cyc}(x^3+3x^2y+3x^2z+2xyz)+\sum_{cyc}3xyz\geq 4\sum_{cyc}(x^2y+x^2z+xyz)$$ or $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is Schur.

Your way is wrong because you took too strong estimation, that got a wrong inequality.

Here happens like the following.

Let we need to prove that $2>1$.

We know that $1<3$, but it does not say that $2>3$ can help because it's just wrong.

The Schur's inequality we can proof so:

Let $x\geq y\geq z.$

Thus, $z(z-x)(z-y)\geq0$ and $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)=\sum_{cyc}x(x-y)(x-z)\geq$$ $$\geq x(x-y)(x-z)+y(y-x)(y-z)=(x-y)^2(x+y-z)\geq0.$$

Answer 2

Use the fact that $ GM \leq AM $. For six numbers $ x_1, x_2, \dots, x_6 $, you have $$ (x_1 x_2 \dots x_6)^{1/6} \leq \frac{x_1 + x_2 + \dots + x_6}{6} $$

Then, you have $$ x y z \leq \frac{x^2 y + x y^2 + y^2 z + y z^2 + z^2 x + z x^2}{6} $$

Answer 3

If we write $x+y+z=1$ (we can do that) then we have to prove $$1+9xyz\geq 4(xy+yz+zx)$$ Clearly one of $x,y,z$ is smaller than $4/9$, we can assume $y<4/9$.

If we put $z=1-x-y$ we get $$0\geq (9y-4)x^2+x(9y^2-13y+4)-(2y-1)^2=:f(x)$$

Since the discirminant for $f(x)$ is: $$\Delta= (9y-4)y(3y-1)^2$$

is also nonegative for all $y<4/9$ we are done.