<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] It's enough to prove our inequality for non-negatives $a$, $b$ and $c$.
Now, by C-S $$\sum_\text{cyc}a\sqrt[4]{\frac{b^4+c^4}{2}}\leq\sum_\text{cyc}a\sqrt{b^2-bc+c^2}\leq\sqrt{\sum_\text{cyc}a\sum_\text{cyc}a(b^2-bc+c^2)}\leq\sum_\text{cyc}a^2$$
because $$\sqrt[4]{\frac{b^4+c^4}{2}}\leq\sqrt{b^2-bc+c^2}$$ it's just $$(b-c)^4\geq0$$ and
$$\sqrt{\sum_\text{cyc}a\sum_\text{cyc}a(b^2-bc+c^2)}\leq\sum_\text{cyc}a^2$$ it's
$$\sum_\text{cyc}(a^4-a^3b-a^3c+a^2bc)\geq0,$$ which is Schur.
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