By Cauchy-Schwarz's inequality, we have $$\sum_{cyc}\frac{x}{\sqrt{\sqrt[4]{y}+\sqrt[4]{z}}}\ge \frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{\sqrt{\sqrt[4]{y}+\sqrt[4]{z}} +\sqrt{\sqrt[4]{z}+\sqrt[4]{x}}+ \sqrt{\sqrt[4]{x}+\sqrt[4]{y}}}.$$So, it remains to prove that $\sum_{cyc}\sqrt{\sqrt[4]{x}+\sqrt[4]{y}} \le \sqrt[4]{108\sum_{cyc}\sqrt{x}}$. By Power-mean inequality, we have $$\sum_{cyc} \sqrt{\sqrt[4]{x}+{\sqrt[4]{y}}} \le \sqrt{6\sum_{cyc}\sqrt[4]{x}} \le \sqrt[4]{108\sum_{cyc}\sqrt{x}}$$as we want.

Here are few lemmas which explain the steps I take 1)$\sum\sqrt{a+b}\leq\sqrt{6(a+b+c)}$ This follows directly by QM-AM 2)$3(a^2+b^2+c^2)\geq (a+b+c)^2$ Trivial after expansion, well known too 3)Titu Lemma Also well known

Shouldn't it be $3$ instead of $\sqrt{2\sqrt{27}}$, because with $\sqrt{2\sqrt{27}}$, we can't reach the equality?

Let $x=u^4,y=v^4,z=w^4$, then by Titu's Lemma, \[\sum_{cyc}\frac{u^4}{\sqrt{v+w}}\geq \frac{(u^2+v^2+w^2)^2}{\sum_{cyc}\sqrt{v+w}}.\]Now it is suffices to prove that \[\frac{(u^2+v^2+w^2)^2}{\sum_{cyc}\sqrt{v+w}}\geq \frac{\sqrt[4]{(u^2+v^2+w^2)^7}}{\sqrt{2\sqrt{27}}}.\]Taking both sides to the power of $4$, we have to prove \[108(u^2+v^2+w^2)\geq \left(\sum_{cyc} \sqrt{v+w}\right)^4\]which by Cauchy we have \[\left(\sum_{cyc}u+v\right)\left(\sum_{cyc}1\right)\geq \left(\sum_{cyc}\sqrt{u+v}\right)^2\]so we are left to prove \[108(u^2+v^2+w^2)\geq 36(u+v+w)^2\]which is just \[u^2+v^2+w^2\geq uv+vw+wu\]by AM-GM. Equality holds at $x=y=z=9$.$\square$

I think that this problem was in 2018 JBMO shortlist

Let $x=a^4,y=b^4,z=c^4$. Then the inequality rewrites as $$\sum_{cyc} \frac{a^4}{\sqrt{b+c}} \geq \frac{\sqrt[4]{(a^2+b^2+c^2)^7}}{\sqrt{2\sqrt{27}}}$$By Titu $$\sum_{cyc} \frac{a^4}{\sqrt{b+c}} \geq \frac{(a^2+b^2+c^2)^2}{\sum_{cyc} \sqrt{b+c}}$$Now it suffices to prove that $$\frac{(a^2+b^2+c^2)^2}{\sum_{cyc} \sqrt{b+c}} \geq \frac{\sqrt[4]{(a^2+b^2+c^2)^7}}{\sqrt{2\sqrt{27}}}$$Raising to the fourth power and cancelling the terms, it's equivalent to $$108(a^2+b^2+c^2) \geq (\sqrt{b+c}+\sqrt{c+a}+\sqrt{a+b})^4$$By C-S $$(\sqrt{b+c}+\sqrt{c+a}+\sqrt{a+b})^2 \leq 6(a+b+c)$$So $$(\sqrt{b+c}+\sqrt{c+a}+\sqrt{a+b})^4 \leq 36(a+b+c)^2$$Now it's enough to prove that $$36(a+b+c)^2 \leq 108(a^2+b^2+c^2)$$$$(a+b+c)^2 \leq 3(a^2+b^2+c^2)$$$$bc+ca+ab \leq a^2+b^2+c^2$$Which is obvious, so we're done.

Steve12345 wrote: Let $x,y,z$ be positive real numbers . Prove: $\frac{x}{\sqrt{\sqrt[4]{y}+\sqrt[4]{z}}}+\frac{y}{\sqrt{\sqrt[4]{z}+\sqrt[4]{x}}}+\frac{z}{\sqrt{\sqrt[4]{x}+\sqrt[4]{y}}}\geq \frac{\sqrt[4]{(\sqrt{x}+\sqrt{y}+\sqrt{z})^7}}{\sqrt{2\sqrt{27}}}$

Notice that the inequality is homogeneous. We will set $\sqrt{x}+\sqrt{y}+\sqrt{z}=1$(the motivation for this is that I want to use Titu and the fact that the RHS already has this term). Applying Titu and simplifying, we find that it is sufficient to prove that $\frac{1}{\sum \sqrt{\sqrt[4]{x}+\sqrt[4]{y}}} \geq \frac{1}{\sqrt{6\sqrt{3}}}\implies \sum \sqrt{\sqrt[4]{x}+\sqrt[4]{y}}\leq \frac{1}{\sqrt{6\sqrt{3}}}$ We want to get rid of the square roots, and the easiest way to do so is C-S. It is easy to finish now. We have that $$(\sum \sqrt[4]{x}+\sqrt[4]{y})(3)\geq (\sum \sqrt{\sqrt[4]{x}+\sqrt[4]{y}})^2,$$ and that $$(\sum \sqrt{x})(3)\geq (\sum \sqrt[4]x)^2\implies \sum \sqrt[4]{x}\leq \sqrt{3},$$so we have that $$\sum \sqrt{\sqrt[4]{x}+\sqrt[4]{y}}\leq \sqrt{(2\sum \sqrt[4]{x})\cdot 3}\leq \sqrt{3}\sqrt{2\sqrt{3}}$$and we are done.

Let $x=a^4,y=b^4,z=c^4.$ The inequality can be rewritten as $$\sum_{cyc}\frac{a^4}{\sqrt{b+c}}\geq\frac{(a^2+b^2+c^2)^{7/4}}{\sqrt{2\sqrt{27}}}.$$By Titu's Lemma, we have $$\sum_{cyc}\frac{a^4}{\sqrt{b+c}}\geq\frac{(a^2+b^2+c^2)^2}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}},$$so it suffices to show that $$\frac{(a^2+b^2+c^2)^2}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}}\geq \frac{(a^2+b^2+c^2)^{7/4}}{\sqrt{2\sqrt{27}}},$$which can be rewritten as $$(a^2+b^2+c^2)^{1/4}\sqrt{2\sqrt{27}}\geq \sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}.$$Let $f(x)=\sqrt{x}$ which is concave. By Jensen's, $$\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\leq 3\sqrt{\frac{2}{3}(a+b+c)}.$$Furthermore, by Power Mean, $$3\sqrt{\frac{2}{3}(a+b+c)}\leq 3\sqrt{\frac{2}{3}(3\sqrt{\frac{a^2+b^2+c^2}{3}})}.$$We can clean up the RHS as $$(a^2+b^2+c^2)^{1/4}\sqrt{2\sqrt{27}},$$so we are done.