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