Nesbitt's Inequality for 4 Variables

Since we have $(x-y)^2\ge 0$, we have, for $x\gt 0,y\gt 0$, $$\begin{align}(x-y)^2\ge 0&\Rightarrow x^2+y^2+2xy\ge 4xy\\&\Rightarrow y(x+y)+x(x+y)\ge 4xy\\&\Rightarrow \frac{1}{x}+\frac 1y\ge\frac{4}{x+y}\end{align}$$ Now set $x=b+c,y=a+d$ and $x=c+d,y=a+b$ to get $$\frac{1}{b+c}+\frac{1}{a+d}\ge\frac{4}{b+c+a+d}$$and$$\frac{1}{c+d}+\frac{1}{a+b}\ge\frac{4}{c+d+a+b}.$$

The simplest is to use the AM-HM inequality; which is a consequence of the AM-GM inequality, but can be proved independently.

Remember the harmonic mean of $x$ and $y$ is the number whose inverse is the arithmetic mean of the inverses of $x$ and $y$. Explicitly: $$\frac1H=\frac12\Bigl(\frac1x+\frac1y\Bigr)\ge \frac1A=\frac2{x+y}$$ Thus $\,\dfrac1x+\dfrac1y\ge\dfrac4{x+y}$.

Apply this inequality twice in $$(a+c)\Bigl(\frac1{b+c}+\frac1{a+d}\Bigl)+(b+d)\Bigl(\frac1{c+d}+\frac1{a+b}\Bigl). $$