Prove that $3(a^5b+b^5c+c^5a)\geq(a^2c+b^2a+c^2b)^2$ 
Let $\sqrt{a}$, $\sqrt{b}$ and $\sqrt{c}$ be sides-lengths of a triangle. Prove that:
$$3(a^5b+b^5c+c^5a)\geq(a^2c+b^2a+c^2b)^2$$

I tried to apply the  way like the proof of the following inequality, but without success.
Let $\sqrt{a}$, $\sqrt{b}$ and $\sqrt{c}$ be sides-lengths of a triangle. Prove that:
$$2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq3+\frac{a}{c}+\frac{b}{a}+\frac{c}{b}.$$
Proof.
We need to prove that
$$\sum_{cyc}(2a^2c-a^2b-abc)\geq0$$ or
$$\sum_{cyc}(a^3-abc)-\sum_{cyc}(b^3-2ab^2+a^2b)\geq0$$ or
$$\frac{a+b+c}{2}\sum_{cyc}(a-b)^2-\sum_{cyc}b(a-b)^2\geq0$$ or
$$\sum_{cyc}(a-b)^2(a+c-b)\geq0.$$
Lemma.
Let $x+y+z\geq0$ and $xy+xz+yz\geq0$. Prove that:
$$(b-c)^2x+(a-c)^2y+(b-c)^2z\geq0.$$
Proof of the lemma.
Let $x+y\geq0$.
If $x+y=0$ then $xy+xz+yz=-x^2\geq0$,
which gives $x=y=0$ and $z\geq0$, which says that
$(b-c)^2x+(a-c)^2y+(a-b)^2z\geq0$ is true.
Thus, we can assume that $x+y>0$ and we need to prove that
$$(b-c)^2x+(a-b+b-c)^2y+(a-b)^2z\geq0$$ or
$$(x+y)(b-c)^2+2y(a-b)(b-c)+(y+z)(a-b)^2\geq0,$$
for which it's enough to prove that
$$y^2-(x+y)(y+z)\leq0$$ or
$$xy+xz+yz\geq0,$$
which ends a proof of the lemma.
Now, $\sum\limits_{cyc}(a+c-b)=a+b+c>0$ and
$$\sum_{cyc}(a+b-c)(a+c-b)=\sum_{cyc}(2ab-a^2)=$$
$$=(\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{b}+\sqrt{c}-\sqrt{a})>0,$$
which ends the proof by the lemma.
Thank you!
 A: $$3(a^5b + b^5c + c^5a) \ge (a^2b + b^2c + c^2a)^2$$
$$\iff 3(abc)^2\left[\left(\frac{a}{b}\right)^3 \cdot \left(\frac{b}{c}\right)^2 + \left(\frac{b}{c}\right)^3 \cdot \left(\frac{c}{a}\right)^2 + \left(\frac{c}{a}\right)^3 \cdot \left(\frac{a}{b}\right)^2\right] \ge (abc)^2\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\right)^2$$
Let $\dfrac{a}{b} = x, \dfrac{b}{c} = y, \dfrac{c}{a} = z \implies xyz = 1$.
$$\iff 3\left(\frac{x^2y}{z} + \frac{y^2z}{x} + \frac{z^2x}{y}\right) \ge (x + y + z)^2$$
, which can be proven by using the Chebyshev inequality, the AM-GM inequality and the Cauchy inequality (which applications can be used in the order listed).
A: Remark: Since no complete solution is available, I give the solution by the Buffalo Way. Hope to see nice solutions.
Let $a = (u+v)^2, b = (v+w)^2, c = (w+u)^2$ for $u, v, w > 0$ (Ravi Substitution). It suffices to prove that
$g(u,v,w) \ge 0$ where $g(u,v,w)$ is a homogeneous polynomial of degree $12$.
WLOG, assume that $w = \min(u, v, w)$. Let $v = w + s$ and $u = w + t$ for $s, t\ge 0$.
It suffices to prove that
$$q_{10}w^{10} + q_9w^9 + q_8w^8 + q_7w^7 + q_6w^6 + q_5w^5 + q_4w^4 + q_3w^3 + q_2w^2 + q_1w + q_0 \ge 0.$$
It is not hard to prove that $q_{10}, q_9, \cdots, q_0\ge 0$.
I do not give the expressions of $q_i, \forall i$. If someone uses the Computer Algebra Systems (CAS), it is an easy job.
