YanYau wrote: Let $a,b,c$ be positive real numbers satisfying $abc=1$. Determine the smallest possible value of $$\frac{a^3+8}{a^3(b+c)}+\frac{b^3+8}{b^3(a+c)}+\frac{c^3+8}{c^3(b+a)}$$ $\frac{a^3+8}{a^3(b+c)}+\frac{b^3+8}{b^3(a+c)}+\frac{c^3+8}{c^3(b+a)}\geq\frac{27}{2}.$ $\sum_{cyc}\frac{1}{a^3(b+c)}\geq\frac{3}{2}\geq\sum_{cyc}\frac{1}{b+c}$

YanYau wrote: Let $a,b,c$ be positive real numbers satisfying $abc=1$. Determine the smallest possible value of $$\frac{a^3+8}{a^3(b+c)}+\frac{b^3+8}{b^3(a+c)}+\frac{c^3+8}{c^3(b+a)}$$ By $AM-GM$ $a^3+1+1 \geq 3a$ and two known results: $\frac{1}{a^2(b+c)}+\frac{1}{b^2(a+c)}+\frac{1}{c^2(a+b)}\geq \frac{3}{2}$ and $\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\geq \frac{3}{2}$

Wangzu wrote: YanYau wrote: Let $a,b,c$ be positive real numbers satisfying $abc=1$. Determine the smallest possible value of $$\frac{a^3+8}{a^3(b+c)}+\frac{b^3+8}{b^3(a+c)}+\frac{c^3+8}{c^3(b+a)}$$ By $AM-GM$ $a^3+1+1 \geq 3a$ and two known results: $\frac{1}{a^2(b+c)}+\frac{1}{b^2(a+c)}+\frac{1}{c^2(a+b)}\geq \frac{3}{2}$ and $\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\geq \frac{3}{2}$ $\sum_{cyc}\frac{a^3+8}{a^3(b+c)}\geq\sum_{cyc}\frac{3a+6}{a^3(b+c)}\geq\frac{3(\sum_{cyc}\frac{1}{a})^2}{2\sum_{cyc}a}+\frac{6(\sum_{cyc}\frac{1}{a})^2}{2\sum_{cyc}bc}$ $=\frac{3(\sum_{cyc}bc)^2}{2\sum_{cyc}a}+3\sum_{cyc}bc\geq \frac{9abc}{2}+3\sum_{cyc}bc\geq\frac{27}{2}. $ Equality holds when $a=b=c=1.$

Let $a,b,c$ be positive real numbers satisfying $abc=1$. Then$$\frac{a^3+\frac{1}{5}}{a^3(b+c)}+\frac{b^3+\frac{1}{5}}{b^3(a+c)}+\frac{c^3+\frac{1}{5}}{c^3(b+a)}\geq \frac{9}{5}.$$Find the minimum value of $k$ for which for all positive real numbers $a, b, c$ satisfying $abc=1$ ,we have$$\frac{a^3+k}{a^3(b+c)}+\frac{b^3+k}{b^3(a+c)}+\frac{c^3+k}{c^3(b+a)}\geq \frac{3(k+1)}{2}.$$

YanYau wrote: Let $a,b,c$ be positive real numbers satisfying $abc=1$. Determine the smallest possible value of $$\frac{a^3+8}{a^3(b+c)}+\frac{b^3+8}{b^3(a+c)}+\frac{c^3+8}{c^3(b+a)}$$ $\sum_{cyc}\frac{a^3+8}{a^3(b+c)}=\sum_{cyc}\frac{1}{b+c}+8\sum_{cyc}\frac{(bc)^2}{a(b+c)}\geq\frac{9}{2\sum_{cyc}a}+\frac{8(\sum_{cyc}bc)^2}{2\sum_{cyc}bc}\geq\frac{9}{2\sum_{cyc}a}+\sum_{cyc}bc+9$ $\geq\frac{9}{2\sum_{cyc}a}+\sqrt{3abc\sum_{cyc}a}+9=\frac{1}{2}\left(\frac{9}{\sum_{cyc}a}+ \sqrt{3\sum_{cyc}a}+ \sqrt{3\sum_{cyc}a}\right)+9\geq \frac{9}{2}+9=\frac{27}{2}. $ Equality holds when $a=b=c=1.$

For me I use Lemmas Cauchy-schwarz

we can use vornicu schur inequality this problem is quite simple by using vornicu schue inequality