Question

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 : \(\dfrac{1}{3a+2b+c}+\dfrac{1}{a+3b+2c}+\dfrac{1}{2a+b+3c}\le\dfrac{8}{3}\)

Answer 1

b) \(\dfrac{1}{3a+2b+c}\le\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{1}{36}\left(\dfrac{3}{a}+\dfrac{2}{b}+\dfrac{1}{c}\right)\)

Tương tự cho 2 cái kia rồi cộng lại

\(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{6}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}.16=\dfrac{8}{3}\)

Đẳng thức xảy ra \(\Leftrightarrow\) ... \(\Leftrightarrow a=b=c=\dfrac{3}{16}\)