My solution, but looking forward to other solutions By AM-GM, $$\frac{a^2}{(b+c)^3}+\frac{9(b+c)}{16}\geq\frac{3}{2}\frac{a}{b+c}$$Summing similar results, we get $$LHS\geq\frac32{\sum\frac{a}{b+c}}-\frac{9}{8}\geq \frac98$$Where the last step follows from Nesbitt

$$\sum_{cyc} \frac{a^2}{(b+c)^3} = \sum_{cyc} \frac{a^2(a+b+c)}{(b+c)^3} = \sum_{cyc} \frac{a^3}{(b+c)^3} + \sum_{cyc} \frac{a^2}{(b+c)^2} \geq \frac{(\sum_{cyc} \frac{a} {b+c})^3}{9} +\frac{(\sum_{cyc} \frac{a} {b+c})^2}{3} \geq \frac{9}{8}$$

Math-wiz wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac{a^2}{(b+c)^3}+\frac{b^2}{(c+a)^3}+\frac{c^2}{(a+b)^3}\geq \frac98$$ From the Generalized Titu Inequality, we have \begin{align*}\frac{a^2}{(b+c)^3}+\frac{b^2}{(c+a)^3}+\frac{c^2}{(a+b)^3} & \geq 3^{1+3-2} \cdot \frac{(a+b+c)^2}{(2(a+b+c))^3} \\& \geq \frac98 \cdot\frac{1}{a+b+c} = \frac98,\end{align*}which thus completes the problem. $\blacksquare$

Drakir wrote: $$\sum_{cyc} \frac{a^2}{(b+c)^3} = \sum_{cyc} \frac{a^2(a+b+c)}{(b+c)^3} = \sum_{cyc} \frac{a^3}{(b+c)^3} + \sum_{cyc} \frac{a^2}{(b+c)^2} \geq \frac{(\sum_{cyc} \frac{a} {b+c})^3}{9} +\frac{(\sum_{cyc} \frac{a} {b+c})^2}{3} \geq \frac{9}{8}$$ Total beauty of homogenisation

Jensen Solves it, as does the Tangent Line Method

My Solution: It's clear,that $a,b,c\in (0,1)$. Write $b+c$ as $1-c$,then our expression is identically as same as: $\sum\frac{a^{2}}{(1-a)^{3}}\geq\frac{9}{8}$ Now consider function $f(x)=\frac{x^{2}}{(1-x)^{3}}$.After some boring calculations,we get that: $f"(x)=\frac{12x^{2}}{(1-a)^{5}}+\frac{12x}{(1-x)^{4}}+\frac{2}{(1-x)^{3}}>0$ when $x\in (0,1)$,so $f$ is convex on $(0,1)$. And using Jensen's Inequality,we get: $f(a)+f(b)+f(c)\geq 3f(\frac{a+b+c}{3})=3\times \frac{3}{8}=\frac{9}{8}$ $Q.E.D$

What are all these overkills guys? Do this. Observe by Cauchy-Schwarz, $$ (b+c+c+a+a+b)\left(\frac{a^2}{(b+c)^3}+\frac{b^2}{(c+a)^3}+\frac{c^2}{(a+b)^3}\right)\geqslant \left(\sum_{{\rm cyc}}\frac{a}{b+c}\right)^2. $$Now use $a+b+c=1$, and the fact that $\sum_{{\rm cyc}} \frac{a}{b+c}\geqslant \frac32$ (well-known) to conclude.

Notice that since $a+b+c=1$ we can rewrite the inequality as $\sum_{cyc}\frac{a^2}{(b+c)^3}=\sum_{cyc}\frac{a^2}{(1-a)^3}$ Now let $f(x)=\frac{x^2}{(1-x)^3}$, furthermore $f'(x)=\frac{2x+x^2}{(1-x)^4}\text{ and }f''(x)=\frac{2(x^2+4x+1)}{(1-x)^5}>0, \forall x\in(0,1)$ therefore $f$ is convex on this interval. $\sum_{cyc}\frac{a^2}{(1-a)^3}=\sum_{cyc}f(a)\overset{\text{Jensen}}{\ge}3f\left(\frac{\sum_{cyc}a}{3}\right)=3f\left(\frac{1}{3}\right)=3\cdot\frac{\frac{1}{9}}{\frac{8}{27}}=3\cdot\frac{3}{8}=\frac{9}{8}$ $\blacksquare$.

By Holder $(1+1+1)(a+b+c)(\frac{a^2}{(b+c)^3}+\frac{b^2}{(c+a)^3}+\frac{c^2}{(a+b)^3}) \geq (\sum{\frac{a}{b+c}})^3 \geq \frac{27}{8}$ By dividing both sides by $3$, we get the desired inequality.

NJOY wrote: Math-wiz wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac{a^2}{(b+c)^3}+\frac{b^2}{(c+a)^3}+\frac{c^2}{(a+b)^3}\geq \frac98$$ From the Generalized Titu Inequality, we have \begin{align*}\frac{a^2}{(b+c)^3}+\frac{b^2}{(c+a)^3}+\frac{c^2}{(a+b)^3} & \geq 3^{1+3-2} \cdot \frac{(a+b+c)^2}{(2(a+b+c))^3} \\& \geq \frac98 \cdot\frac{1}{a+b+c} = \frac98,\end{align*}which thus completes the problem. $\blacksquare$ This is a wrong application of Generalized Titu because of $k_{1}\leq k_{2}+1$ mentioned in the link you shared. Gives true answer however this is a wrong solution.

Instead multiplicate both numerator-denominator with $a^3$ for substituting truely Titu $$\frac{a^2}{(b+c)^3}+\frac{b^2}{(c+a)^3}+\frac{c^2}{(a+b)^3}=\sum_{cyc}{\frac{a^5}{(ab+ac)^3}}\geq \frac{(a+b+c)^5}{3.(2(ab+bc+ca))^3}$$$$\geq \frac{1}{3.8.\frac{1}{27}}=\frac{9}{8}$$

Actually