Question

Cho x,y,z >0 và xy\(\ge\)12 ,yz\(\ge8\) CMR:

(x+y+z) +2(\(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\)) +\(\dfrac{8}{xyz}\) \(\ge\dfrac{121}{12}\)

Giải giúp mình với !!!

Answer 1

Dự đoán điểm rơi: x=3 ; y =4;z =2

ÁP dụng AM-Gm ta có:

\(\dfrac{8}{xyz}+\dfrac{x}{9}+\dfrac{y}{12}+\dfrac{z}{6}\ge4\sqrt[4]{\dfrac{8}{9.12.6}}=\dfrac{4}{3}\)

\(\dfrac{2}{xy}+\dfrac{x}{18}+\dfrac{y}{24}\ge3\sqrt[3]{\dfrac{2}{18.24}}=\dfrac{1}{2}\)

\(\dfrac{2}{yz}+\dfrac{y}{16}+\dfrac{z}{8}\ge3\sqrt[3]{\dfrac{2}{16.8}}=\dfrac{3}{4}\)

\(\dfrac{2}{xz}+\dfrac{z}{6}+\dfrac{x}{9}\ge3\sqrt[3]{\dfrac{2}{6.9}}=1\)

\(\dfrac{13}{18}x+\dfrac{13}{24}y\ge2\sqrt{\dfrac{169}{18.24}xy}\ge\dfrac{13}{3}\)

\(\dfrac{13}{24}z+\dfrac{13}{48}y\ge2\sqrt{\dfrac{169}{24.48}.yz}\ge\dfrac{13}{6}\)

Cộng tất cả theo vế ,ta thu được Đpcm.