I have a doubt regarding this inequality-
Suppose $a, b, c$ are positive real numbers with sum $3$. Prove that $$\sqrt{a+\sqrt{b^2+c^2}}+\sqrt{b+\sqrt{c^2+a^2}}+\sqrt{c+\sqrt{a^2+b^2}} \ge 3\sqrt{\sqrt{2}+1}$$
I have a proof of it in my book, which is too long and complicated. I don't doubt it's validity, but accustomed methods are not so helpful, provided the proof has a lot of addition of new expressions.
I wish to have a solution, an easier one, but with commentary (a supplement of words that describe the steps to be more natural, not something which arises out of nowhere. )
Thanks!
By C-S twice we obtain: $$\sum_{cyc}\sqrt{a+\sqrt{b^2+c^2}}=\sqrt{3+\sum_{cyc}\left(\sqrt{a^2+b^2}+2\sqrt{(a+\sqrt{b^2+c^2})(b+\sqrt{a^2+c^2})}\right)}\geq$$ $$\geq\sqrt{3+\sum_{cyc}\left(\sqrt{a^2+b^2}+2\sqrt{\left(a+\frac{b+c}{\sqrt2}\right)\left(b+\frac{a+c}{\sqrt2}\right)}\right)}=$$ $$=\sqrt{3+\sum_{cyc}\left(\sqrt{a^2+b^2}+\sqrt2\sqrt{(\sqrt2a+b+c)(\sqrt2b+a+c)}\right)}=$$ $$=\sqrt{3+\sum_{cyc}\left(\sqrt{a^2+b^2}+\sqrt2\sqrt{((\sqrt2-1)a+a+b+c)((\sqrt2-1)b+a+b+c)}\right)}\geq$$ $$\geq\sqrt{3+\sum_{cyc}\left(\sqrt{a^2+b^2}+\sqrt2((\sqrt2-1)\sqrt{ab}+3)\right)}=$$ $$=\sqrt{\sum_{cyc}\left(\sqrt{a^2+b^2}+(2-\sqrt2)\sqrt{ab}\right)+3+9\sqrt2}.$$ Id est, it's enough to prove that: $$\sum_{cyc}\left(\sqrt{a^2+b^2}+(2-\sqrt2)\sqrt{ab}\right)+3+9\sqrt2\geq9\sqrt2+9$$ or $$\sum_{cyc}\left(\sqrt{a^2+b^2}+(2-\sqrt2)\sqrt{ab}\right)\geq2(a+b+c)$$ or $$\sum_{cyc}\left(\sqrt{a^2+b^2}+(2-\sqrt2)\sqrt{ab}-a-b\right)\geq0$$ or $$\sum_{cyc}\left(\frac{(a-b)^2}{\sqrt{a^2+b^2}+\sqrt{2ab}}-(\sqrt{a}-\sqrt{b})^2\right)\geq0$$ or $$\sum_{cyc}\frac{(\sqrt{a}-\sqrt{b})^2((\sqrt{a}+\sqrt{b})^2-\sqrt{a^2+b^2}-\sqrt{2ab})}{\sqrt{a^2+b^2}+\sqrt{2ab}}\geq0$$ or $$\sum_{cyc}\frac{(\sqrt{a}-\sqrt{b})^2(a+b-\sqrt{a^2+b^2}+(2-\sqrt2)\sqrt{ab})}{\sqrt{a^2+b^2}+\sqrt{2ab}}\geq0,$$ which is obvious.