If a,b,c,d are four unequal positive number which are in A.P then

b = a + p,c = a + 2p,d = a + 3p
$\begin{array}{l}\frac{\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{d}}}{\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}}}=\frac{(1+\mathrm{p})(1+2\mathrm{p})}{1(\mathrm{a}+3\mathrm{p})}\\ =\frac{{\mathrm{a}}^2+3\mathrm{ap}+2{\mathrm{p}}^2}{{\mathrm{a}}^2+3\mathrm{ap}}>1\\ \therefore \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{d}}>\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}}\\ \left(\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}}\right)(\mathrm{a}+\mathrm{d})=\left(\frac{1}{\mathrm{a}+\mathrm{p}}+\frac{1}{\mathrm{a}+2\mathrm{p}}\right)(\mathrm{a}+\mathrm{a}+3\mathrm{p})\\ =\frac{(2\mathrm{a}+3\mathrm{p}{)}^2}{{\mathrm{a}}^2+3\mathrm{ap}+2{\mathrm{p}}^2}=4\frac{{\mathrm{p}}^2}{{\mathrm{a}}^2+3\mathrm{ap}+2{\mathrm{p}}^2}>4\end{array}$