Question. Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=\sqrt{a}+\sqrt{b}+\sqrt{c}>0.$ Find the minimal value of $P$ $$P=(a+b)(b+c)(c+a).$$
I check that Mininum is $8$ achieved at $a=b=c=1$ or $a=0; b=c=\sqrt[3]{4}.$
I hope the following will help $$9(a+b)(b+c)(c+a)\ge 8(a+b+c)(ab+bc+ca)=8(a+b+c)(\sqrt{a}+\sqrt{b}+\sqrt{c})$$ The rest is proving $(a+b+c)(\sqrt{a}+\sqrt{b}+\sqrt{c})\ge 9.$ I still don't know how to continue.
Hope you help me. Thank you.
The inequality, which you got is wrong.
Try $(a,b,c)=\left(\sqrt[3]4,\sqrt[3]4,0\right).$
We need to prove that $$2\sqrt[3]4\cdot2\sqrt[6]4\geq9,$$ which is $8\geq9.$
My solution.
We need to prove that $$\prod_{cyc}(a^2+b^2)(a+b+c)^2\geq8(a^2b^2+a^2c^2+b^2c^2)^2,$$ where $a$, $b$ and $c$ are non-negatives.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that: $$((a^2+b^2+c^2)(a^2b^2+a^2c^2+b^2c^2)-3a^2b^2c^2)(a+b+c)^2\geq8(a^2b^2+a^2c^2+b^2c^2)^2$$ or
$$((9u^2-6v^2)(9v^4-6uw^3)-3w^6)\geq8(9v^4-6uw^3)^2$$ and we see that we need to prove that $f(w^3)\geq0,$ where $f$ is a concave function.
Thus, by $uvw$ it's enough to prove $f(w^3)\geq0$ for equality case of two variables.
For $b=c=0$ we obtain an equality and $b=c=1$ gives $$4(a^2+1)^2(a+2)^2\geq8(2a^2+1)^2$$ or $$a(a-1)^2(a^3+6a^2+a+4)\geq0,$$ which ends the proof.