Quidditch wrote: Let $a,b,c,x,y$ be positive real numbers such that $abc=1$. Prove that $$\frac{a^5}{xc+yb}+\frac{b^5}{xa+yc}+\frac{c^5}{xb+ya}\geq \frac{9}{(x+y)(a^2+b^2+c^2)}.$$ https://artofproblemsolving.com/community/c6h480595p2691557

$\frac{a^5}{xc+yb}+\frac{b^5}{xa+yc}+\frac{c^5}{xb+ya} \geq \frac{(a^3+b^3+c^3)^2}{(x+y)(ab+bc+ca)} \geq \frac{9}{(x+y)(ab+bc+ca)} \geq \frac{9}{(x+y)(a^2+b^2+c^2)}$

Alternative solution: $$\frac{a^5}{xc+yb}+\frac{b^5}{xa+yc}+\frac{c^5}{xb+ya} \geq\frac{(\sqrt{a^5}+\sqrt{b^5}+\sqrt{c^5})^2}{(x+y)(a+b+c)}\geq\frac{9}{(x+y)(a+b+c)} $$ We want $a^2+b^2+c^2\geq a+b+c$, by AM-GM, $4a^2+b^2+c^2\geq 6a$, $a^2+4b^2+c^2\geq 6b$, $a^2+b^2+4c^2\geq 6c$.

Notice that $\sum_{cyc}\frac{a^5}{xc+yb}\overset{\text{Generalized T2'S}}{\ge}\frac{\left(\sum_{cyc}a\right)^5}{3^3\cdot(x+y)\sum_{cyc}a}$ Thus the inequality transforms into $\frac{\left(\sum_{cyc}a\right)^5}{3^3\cdot(x+y)\sum_{cyc}a}\ge\frac{9}{(x+y)\sum_{cyc}a^2}\Longleftrightarrow\frac{\left(\sum_{cyc}a\right)^4}{3^3}\ge\frac{9}{\sum_{cyc}a^2}\Longrightarrow\left(\sum_{cyc}a\right)^4\sum_{cyc}a^2\ge3^5$ Which is clearly true as $\sum_{cyc}a\overset{\text{AM-GM}}{\ge}3\sqrt[3]{abc}=3\Longrightarrow\left(\sum_{cyc}a\right)^4\ge3^4\text{ and }\sum_{cyc}a^2\ge3\sqrt[3]{a^2b^2c^2}=3$ and by multiplying the inequalities together we get the desired result $\blacksquare$.

pretty much the same as @qinghong's solution