A cyclic, non-symmetric inequality

Question

If $x,y,z\geq 0$, prove $$ (x^2+y^2+z^2)^2\geq 3(x^3y+y^3z+z^3x) $$ With these nonsymmetric inequalities the usual tools like Muirhaed or Schur do not apply, and also AM-GM doesn't seem to be of any help. Also, it's non-factorizable.

Answer 1

Because $$(x^2+y^2+z^2)^2-3(x^3y+y^3z+z^3x)=\frac{1}{2}\sum\limits_{cyc}(x^2-z^2-2xy+xz+yz)^2\geq0$$ Also we have $$(x^2+y^2+z^2)^2-3(x^3y+y^3z+z^3x)=\frac{1}{6}\sum\limits_{cyc}(x^2+y^2-2z^2-3xy+3xz)^2\geq0$$

Answer 2

I found a general solution a few years ago. Here it is

Consider the 4-degree polynomial:

$$ \sum_{cyc} a^4 + ( \frac {\Delta_1 + \Delta_2}{2} - 1 ) \sum_{cyc}a^2b^2 + m_1 \sum_{cyc}a^3b + m_2\sum_{cyc}ab^3 + m_3 abc (a + b + c ) $$
where $$ \Delta_1 = m_1^2 + m_1m_2 + m_2^2 $$

$$ \Delta_2 = m_1 + m_2 + m_3 $$

If $\Delta= \Delta_1 + 3\Delta_2 \geq 0$ then $ P(a,b,c) \geq 0$ for all real numbers $ a,b,c,m_1,m_2,m_3$

Proof: $$ P(a,b,c) =\frac{1}{2} \sum_{cyc}(a^2 - c^2 + pab + qbc + rca)^2$$

where $p,q,r$ are roots of the polynomial

$$ x^3-\sqrt{\Delta}x^2+\Delta_2 x-\frac{+(2m_1+m_2+\sqrt{\Delta})(-m_1-2m_2+\sqrt{\Delta})(-m_1+m_2+\sqrt{\Delta}+)}{27} =0 $$
also $p- q = m_1+m_2$ and $q - r =-m_2$

You can apply this formula to your problem. Good luck !

P/S: Here is a nice reference for your problem: Ref 1