<s>[INST] Hard and non-trivial inequality with three variable reals For all reals $a$, $b$, $c$, show that $$a^2+b^2+c^2 \geq a\sqrt[\leftroot{-1}\uproot{1}4]{\frac{b^4+c^4}{2}} + b\sqrt[\leftroot{-1}\uproot{1}4]{\frac{c^4+a^4}{2}} + c\sqrt[\leftroot{-1}\uproot{1}4]{\frac{a^4+b^4}{2}}.$$
I tried to use Holder inequality:
$$\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)(a^2+b^2+c^2)(a^2+b^2+c^2)\Big(b^4+c^4+c^4+a^4+a^4+b^4\Big) \geq \text{RHS}^4$$
or
$$(a^2+b^2+c^2)^4 \geq (1+1+1)(a+b+c)^2 \Big(\sum_\text{cyc}{\frac{a^2(b^4+c^4)}{2}} \Big) \geq \text{RHS}^4$$
But got stuck after that (the $\geq$ is reversed).
 [/INST] Your second attempt gives another solution!
Indeed, for non-negative variables we need to prove that:
$$2(a^2+b^2+c^2)^4\geq3(a+b+c)^2\sum_{cyc}(a^4b^2+a^4c^2)$$ or
$$2(a^2+b^2+c^2)^4\geq3(a+b+c)^2\left(\sum_{cyc}a^2\sum_{cyc}a^2b^2-3a^2b^2c^2\right).$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$2(9u^2-6v^2)^4\geq27u^2((9u^2-6v^2)(9v^4-6uw^3)-3w^6)$$ or $f(w^3)\geq0,$ where
$$f(w^3)=u^2w^6+6u^3(3u^2-2v^2)w^3+2(3u^2-2v^2)^4-9u^2v^4(3u^2-2v^2).$$
But it's obvious that $f$ increases, which says that it's enough to prove our inequality for the minimal value of $w^3$, which by $uvw$(about $uvw$ see here: https://artofproblemsolving.com/community/c6h278791 )
happens in the following cases:

*

*$w^3=0.$
Let $c=0$ and $b=1$.
We need to prove: $$2(a^2+1)^4\geq3(a+1)^2(a^4+a^2),$$ which is true by C-S and AM-GM:
$$2(a^2+1)^4=2(a^2+1)(a^2+1)^2(a^2+1)\geq(a+1)^2\cdot4a^2(a^2+1)\geq3(a+1)^2(a^4+a^2);$$
2. Two variables are equal.
Let $b=c=1$.
We need to prove that:
$$2(a^2+2)^4-3(a+2)^2(2a^4+2a^2+2)\geq0$$ or $$(a-1)^2(a^6+2a^5+8a^4+2a^3+5a^2-4a+4)\geq0$$ and we are done.
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