Doubt in solution of problem in secrets in inequalities by pham kim hung Question -
Suppose that $a, b, c$ are three side-lengths of a triangle with perimeter
3. Prove that
$$
\frac{1}{\sqrt{a+b-c}}+\frac{1}{\sqrt{b+c-a}}+\frac{1}{\sqrt{c+a-b}} \geq \frac{9}{a b+b c+c a}
$$
Solution. Let $x=\sqrt{b+c-a}, y=\sqrt{c+a-b}, z=\sqrt{a+b-c} .$ We get $x^{2}+y^{2}+z^{2}=$
3. The inequality becomes
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{36}{9+x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}
$$
Let $m=x y, n=y z, p=z x .$ The inequality above is equivalent to
$$
(m+n+p)\left(m^{2}+n^{2}+p^{2}+9\right) \geq 36 \sqrt{m n p}
$$
which is obvious by AM-GM because
$$
m+n+p \geq 3 \sqrt[3]{m n p},   m^{2}+n^{2}+p^{2}+9 \geq 12 \sqrt[6]{m n p}
$$
now i did not understand how they got this last inequality $m^{2}+n^{2}+p^{2}+9 \geq 12 \sqrt[6]{m n p}
$
by am-gm ...
thankyou

Hint: Write $9=1+1+\cdots +1$ and then apply AM-GM
$$m^2+n^2+p^2+1+\cdots +1\geq 12\sqrt[12]{m^2n^2p^2}=12\sqrt[6]{mnp}$$
