Vì \(\frac{a+b+c+d}{ab}+\frac{a+b+c+d}{ac}+\frac{a+b+c+d}{ad}\)
\(=\frac{a+b}{ab}+\frac{c+d}{ab}+\frac{a+b}{ac}+\frac{a+b}{ad}+\frac{c+d}{ac}+\frac{c+d}{ad}\)
\(=\left(a+b\right)\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}\right)+\left(d+c\right)\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}\right)\)
Áp dụng bất đẳng thức:
\(\left(a+b\right)\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}\right)\ge18\)
\(\left(c+d\right)\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}\right)\ge18\)
\(\Rightarrow\left(a+b\right)\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}\right)+\left(c+d\right)\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}\right)\ge36\)
\(\Rightarrow\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad}\ge36\left(đpcm\right)\)
