a, cho \(a>0\), \(b>0\) . CM : \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
b , cho 3 số a , b , c thỏa mãn \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=16\)
CM : \(\dfrac{1}{3a+2b+c}+\dfrac{1}{a+3b+2c}+\dfrac{1}{2a+b+3c}\le\dfrac{8}{3}\)