For the non-negative real numbers $a, b, c$ prove that $$(a^2+2)(b^2+2)(c^2+2)\geq 3(a+b+c)^2$$
What I did is applying Holder's inequality in LHS:$$(a^2+(\sqrt{2})^2)(b^2+(\sqrt{2})^2)(c^2+(\sqrt{2})^2) \geq (abc + 2\sqrt{2})^2$$
Then it suffices to prove that $$(abc+2\sqrt2)^2 \geq 3(a+b+c)^2 \\ \Rightarrow abc+2\sqrt2 \geq \sqrt3(a+b+c)$$
But I don't know how to proceed. I also think I applied Holder's Inequality incorrectly.
By C-S $ (a^{2}+2)\left(1+\frac{(b+c)^{2}}{2}\right)\geq(a+b+c)^{2}$.
Hence, it remains to prove that $ (b^{2}+2)(c^{2}+2)\geq3\left(1+\frac{(b+c)^{2}}{2}\right)$,
which is equivalent to $ (b-c)^{2}+2(bc-1)^{2}\geq0$.
Among numbers $a-1, b-1, c-1$ there are two having the same sign, say $a-1, b-1$. In other words, $(a-1)(b-1)\ge 0.$ Multiplying both sides by a positive number $(a+1)(b+1)$ we get \begin{align*} \left(a^2-1\right)\left(b^2-1\right)&\ge 0 \\ a^2b^2-a^2-b^2+1&\ge 0 \\ a^2b^2+2a^2+2b^2+4&\ge 3\left(a^2+b^2+1\right) \\ \left(a^2+2\right)\left(b^2+2\right) &\ge 3\left(a^2+b^2+1\right) \\ \left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right) &\ge 3\left(a^2+b^2+1\right)\left(c^2+2\right). \end{align*}
Using Schwarz inequality we find $$\left(a^2+b^2+1\right)\left(c^2+2\right)=\left(a^2+b^2+1\right)\left(1+1+c^2\right) \ge (a+b+c)^2.$$
Therefore $$\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right) \ge 3\left(a^2+b^2+1\right)\left(c^2+2\right) \ge 3(a+b+c)^2.$$
