$\frac{1}{{{K_R}}} = \frac{1}{{{K_1}}} + \frac{1}{{{K_2}}};If\,{K_1} = {K_2} = k$

$\frac{1}{{{k_R}}} = \frac{1}{K} + \frac{1}{K} = \frac{2}{K} \Rightarrow {K_R} = \frac{K}{2}$

Which is less than $K$.

$If\,{K_1} > {K_2}\,suppose\,{K_1} = {K_2} + x$

$\frac{1}{K} = \frac{1}{{{K_1}}} + \frac{1}{{{K_2}}} = \frac{{{K_2} + {K_1}}}{{{K_1}{K_2}}}$

$ \Rightarrow \frac{1}{K} = \frac{{{K_2} + {K_2} + x}}{{\left( {{K_2} + x} \right){K_2}}} \Rightarrow K = \frac{{K_2^2 + {K_2}x}}{{2{K_2} + x}}$

$Now,\,{K_2} - K = {K_2} - \frac{{K_2^2 + {K_2}x}}{{2{K_2} + x}}$

$ = \frac{{2K_2^2 + {K_2}x - K_2^2 - {K_2}x}}{{\left( {2{K_2} + x} \right)}}$

$ = \frac{{K_2^2}}{{2{K_2} + x}} = positive$

$So.{K_2} > K,so\,the\,value\,of\,K\,is\,smaller\,than\,$

${K_2}\,and\,{K_1}.$