Algebraic inequality $\sum \frac{x^3}{(x+y)(x+z)(x+t)}\geq \frac{1}{2}$

This is a solution using Carlson inequality: if $x_{ij}(1\leq i\leq m,1\leq j\leq n)$ are non-negative reals, then $$ \prod_{s=1}^m\sum_{t=1}^nx_{st}\geq\left(\sum_{t=1}^n\left(\prod_{s=1}^mx_{st}\right)^{\frac{1}{m}}\right)^{m} $$

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Proof. According to the Carlson inequality, $$ \begin{align} &\sum\frac{x^3}{(x+y)(x+z)(x+t)}\cdot\sum(x+y)\cdot\sum(x+z)(x+t)\\ \geq&\left(\sum x\right)^3\\ \because&\sum(x+y)=2(x+y+z+t),\sum(x+z)(x+t)=(x+y+z+t)^2\\ \therefore&\sum\frac{x^3}{(x+y)(x+z)(x+t)}\cdot2(x+y+z+t)\cdot(x+y+z+t)^2\geq(x+y+z+t)^3 \end{align} $$ One can now easily observe that the original inequality is true.

As inequality homogenous WLOG $x+y+z+t=4$

By AM-GM$$\sum \frac{x^3}{(x+y)(x+t)(x+z)} \ge\sum \frac{27x^3}{{(3x+y+z+t)}^3}= \sum \frac{27x^3}{{(2x+4)}^3}$$ By tangent line method $$\frac{x^{3}}{\left(2x+4\right)^{3}}\ge \frac{1}{6^{3}}+\frac{1}{108}\left(x-1\right)$$ $$\iff -\frac{\left(x-1\right)^{2}\left(x^{2}-6x-4\right)}{108\left(x+2\right)^{3}}\ge 0$$ which is true as $x^2-6x-4<0,\forall x\in[0,4]$ So $$ \sum \frac{27x^3}{{(2x+4)}^3}\ge \frac{27 \cdot 4}{6^3}+\frac{0}{108}=\frac{1}{2}$$ done!!