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Solnนี้ credit AOPSครับ
Homogenise this into
\[\sum_{cyc}\frac{a^2}{(a+b)^2+(a+b+c)^2}\leq \frac{3^7}{13}\cdot\frac{a^8+b^8+c^8}{(a+b+c)^6(ab+bc+ca)}.\]
Now we can scrap the previous condition of $a+b+c=1$ and assume that $a+b+c=3.$ Then it suffices to check that
\[\sum_{cyc}\frac{a^2}{(a+b)^2+9}\leq \frac 3{13}\cdot\frac{a^8+b^8+c^8}{(ab+bc+ca)}.\]
Note that we have the following identity: $\frac{9a^2}{(a+b)^2+9}=a^2-\frac{a^2(a+b)^2}{(a+b)^2+9},$
Which, in accordance with the Cauchy-Schwarz inequality, leads us to
\[\sum_{cyc}\frac{9a^2}{(a+b)^2+9}\leq\sum_{cyc}a^2-\frac{\left(\sum a^2+\sum ab\right)^2}{2(\sum a^2+\sum ab)+27}.\]
Let $x=\sum_{cyc}a^2$ and $y=\sum_{cyc}ab,$ so that we have $x+2y=9$ and $x\ge y,$ leading to
\[\frac{(x+y)^2}{2(x+y)+27}=\frac{x+y}{2+\frac{27}{x+y}}\geq \frac{2(x+y)}{13}.\]
So it is sufficient to check that
\[x-\frac{2(x+y)}{13}\leq\frac{27}{13}\cdot\frac{a^8+b^8+c^8}{ab+bc+ca};\]
Which, on using the inequality $27(a^8+b^8+c^8)\geq (a^2+b^2+c^2)^4,$ reduces to
\[\frac{x^4}{y}+2y\geq 11x\]
Now note that, \[\begin{aligned}\frac{x^4}{y}+2y\geq \frac{9x^2}{y}+2y&=\frac{7x^2}{y}+\left(\frac{2x^2}{y}+2y\right)\\&\geq 7x+4x=11x;\end{aligned}\]
Where the last step follows from $x\geq y$ and the AM-GM inequality. Equality holds in the original inequality iff $a=b=c=\frac13.\Box$