Any one?

Hopeooooo wrote: Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=1$ prove that: $\left(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\right)^5 \geq 5^5\left(\frac{ac}{27}\right)^2$ \begin{align*} \left(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\right)^5=\left(\left(\frac{a^2}{a+b}+\frac{c^2}{c+d}\right)+\left(\frac{b^2}{b+c}+\frac{d^2}{d+a}\right)\right)^5\\ \geq\left(\frac{(a+c)^2}{a+b+c+d}+\frac{(b+d)^2}{a+b+c+d}\right)^5=\left(2(a+c)^2-2(a+c)+1\right)^5=\left(\frac{8\left(\frac{3(a+c)}{2}\right)^2-12\left(\frac{3(a+c)}{2}\right)+9}{9}\right)^5\\ \geq\left(\frac{2\left(\frac{3(a+c)}{2}\right)^2+3}{9}\right)^5\geq\left(\frac{1}{9}\cdot5\sqrt[5]{\left(\left(\frac{3(a+c)}{2}\right)^2\right)^2\cdot1^3}\right)^5=5^5\left(\frac{(a+c)^2}{4\cdot3^3}\right)^2\geq5^5\left(\frac{ac}{27}\right)^2 \end{align*}Equality holds for $a=c=\frac{1}{3},b=d=\frac{1}{6}$.