Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c\neq0$. Prove that: $$\sqrt{8a(b+c)+(b-c)^{2}}+\sqrt{8b(c+a)+(c-a)^{2}}+\sqrt{8c(a+b)+(a-b)^{2}}\ge\frac{12(ab+bc+ca)}{a+b+c}.$$
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I solved this problem by the following way.
Let $c=\min\{a,b,c\}$.
We'll prove that $$\sqrt{8a(b+c)+(b-c)^2}+\sqrt{8b(a+c)+(a-c)^2}\geq\sqrt{32ab+16ac+16bc+(a+b-2c)^2}.$$ Indeed, after squaring of the both sides we need to prove $$\sqrt{(8a(b+c)+(b-c)^2)(8b(a+c)+(a-c)^2)}\geq c^2+3ac+3bc+9ab$$ and after squaring again we obtain $$(a-b)^2(ab+ac+bc-3c^2)\geq0,$$ which is obvious.
Thus, it's enough to prove that: $$\sqrt{32ab+16ac+16bc+(a+b-2c)^2}+\sqrt{8c(a+b)+(a-b)^2}\geq\frac{12(ab+ac+bc)}{a+b+c}.$$
Now, let $ab=const$ and $$f(a+b)=(a+b+c)\left(\sqrt{32ab+16ac+16bc+(a+b-2c)^2}+\sqrt{8c(a+b)+(a-b)^2}\right)-12(ab+ac+bc).$$ Thus,$$f'(a+b)\geq\sqrt{32ab+16ac+16bc+(a+b-2c)^2}+\sqrt{8c(a+b)+(a-b)^2}-12c\geq$$ $$\geq\sqrt{32ab+16ac+16bc}+\sqrt{8c(a+b)}-12c\geq0,$$ which says that it's enough to prove $f(a+b)\geq0$ for a minimal value of $a+b$,
which by AM-GM happens for $a=b$.
Since the inequality $f(a+b)\geq0$ is homogeneous, it's enough to assume $a=b=1$ and we need to prove that: $$(c+2)\left(\sqrt{32+32c+(2-2c)^2}+\sqrt{16c}\right)\geq12(2c+1)$$ or $$(c+2)(c+3)+2(c+2)\sqrt{c}\geq6(2c+1)$$ and since by AM-GM $$\sqrt{c}\geq\frac{2c}{c+1},$$ it remains to prove $$(c+2)(c+3)+\frac{4(c+2)c}{c+1}\geq6(2c+1)$$ or $$(c-1)^2c\geq0,$$ which finishes the solution.
I am looking for an alternative solution of this problem.
It seems that the Ji Chen's Lemma may help, but I have no a proof.
Thank you for your interest!