Prove that $3(a^5b+b^5c+c^5a)\geq(a^2c+b^2a+c^2b)^2$

Question

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!

Answer 1

Too long for a comment.

The River Li's solution we can rewrite so.

Let $c=\max\{a,b,c\}$, $a=(x+u)^2,$ $b=(x+v)^2$ and $c=(x+u+v)^2,$ where $x>0$,$u\geq0$ and $v\geq0.$

Thus, $$3(a^5b+b^5c+c^5a)-(a^2c+b^2a+c^2b)^2=$$ $$=60(u^2-uv+v^2)x^{10}+6(58u^3+37u^2v-11uv^2+58v^3)x^9+$$ $$+(959u^4+2078u^3v+1149u^2v^2+350uv^3+959u^4)x^8+$$ $$+8(207u^5+750u^4v+922u^3v^2+428u^2v^3+133uv^4+207u^5)x^7+$$ $$+4(493u^6+2500u^5v+4514u^4v^2+3866u^3v^3+1281u^2v^4+338uv^5+493v^6)x^6+$$ $$+16(105u^7+687u^6v+1602u^5v^2+1936u^4v^3+1183u^3v^4+249u^2v^5+57uv^6+105v^7)x^5+$$ $$+8(129u^8+1035u^7v+2943u^6v^2+4478u^5v^3+3925u^4v^4+1705u^3v^5+150u^2v^6+33uv^7+129v^8)x^4+$$ $$+2(229u^9+2130u^8v+7152u^7v^2+12968u^6v^3+14432u^5v^4+9422u^4v^5+2748u^3v^6-252u^2v^7-27uv^8+225v^9)x^3+$$ $$+(133u^{10}+1438u^9v+5586u^8v^2+11696u^7v^3+15552u^6v^4+13164u^5v^5+6410u^4v^6+928u^3v^7-529u^2v^8-68uv^9+133v^{10})x^2+$$ $$+4(u^2+uv+v^2)(6u^9+66u^8v+247u^7v^2+443u^6v^3+470u^5v^4+271u^4v^5+37u^3v^6-44u^2v^7-11uv^8+6v^9)x+$$ $$+2(u^2+uv+v^2)^3(u^3+5u^2v+2uv^2-v^3)^2\geq0.$$ The inequality in my open post by the same way gives: $$2(a^2c+b^2a+c^2b)-3abc-a^2b-b^2c-c^2a=$$ $$=4(u^2-uv+v^2)x^4+2(6u^3-7u^2v+5uv^2+6v^3)x^3+$$ $$+(13u^4-14u^3v-9u^2v^2+34uv^3+13v^4)x^2+$$ $$+2(3u^5-2u^4v-10u^3v^2+8u^2v^3+13uv^4+3v^5)x+(u^3-3uv^2-v^3)^2\geq0.$$ I think, from the expression $2(u^2+uv+v^2)^3(u^3+5u^2v+2uv^2-v^3)^2$ we can learn that maybe there is a solution for the starting inequality by the lemma.