Prove this inequality: $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$

By AM-GM

$$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \frac{2\sqrt{bc}}{\sqrt{a}}+\frac{\sqrt{ca}}{\sqrt{b}}+\frac{2\sqrt{ab}}{\sqrt{c}} = \frac2a+\frac2b+\frac2c$$

For the last equality $abc=1$ was used.

Then again by AM-GM; $$\frac1a+\frac1b+\frac1c \ge \frac3{\sqrt[3]{abc}}=3$$

and again by a well known inequality $$\frac1a+\frac1b+\frac1c \ge \frac1{\sqrt{bc}}+\frac1{\sqrt{ca}}+\frac1{\sqrt{ab}}=\sqrt{a}+\sqrt{b}+\sqrt{c}$$

For the last equality $abc=1$ was used. Combine the last two inequalities to get the desired result.

You already have a proof using AM-GM and Muirhead. For a pure AM-GM proof, note that $\frac12 \left(\frac{a}b+\frac{b}a\right) \geqslant 1$ and so on, so half the LHS takes care of the $3$ on RHS.

For the rest, note you can cyclically sum the (weighted) AM-GMs: $$\frac5{18} \left(\frac{a}b+\frac{a}c\right)+\frac2{18} \left(\frac{b}a+\frac{b}c\right)+\frac2{18} \left(\frac{c}a+\frac{c}b\right)\geqslant \frac{\sqrt[3]a}{\sqrt[6]{b c}}=\sqrt{a}$$

Id est, we need to prove that $\sum\limits_{cyc}(a^2b+a^2c-a^{\frac{4}{3}}b^{\frac{5}{6}}c^{\frac{5}{6}}-abc)\geq0$

which is AM-GM and Muirhead because $(2,1,0)\succ\left(\frac{4}{3},\frac{5}{6},\frac{5}{6}\right)$

Your another problem can be solved by the same way.