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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.

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

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    There are an infinitely many similar solutions.2017-01-22
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    OK, this works, but these solutions are very non-obvious. How did you even get them?!?2017-01-22
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    @Mak It's a lot of work. Let $\sum\limits_{cyc}(x^4+2x^2y^2-3x^3y)=\sum\limits_{cyc}(kx^2+ny^2+mz^2+pxy+qxz+ryz)^2$ and to solve the system with variables $k$, $n$, $m$, $p$, $q$ and $r$. We have an equality for $x=y=z=1$ and more. In any case, it's a solution of your problem. I have another ways for the proof. Do you want to see them?2017-01-22
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    Yes, please :) This method is tedious, but if there is a solution, we'll get to it, which is nice :)2017-01-23
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    @Maki 2 Solutions, which I posted are the best and short. Another solutions are much harder. I think you'll not like this.2017-01-23
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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