Đáp án: $B\ge 16$
Giải thích các bước giải:
Ta có:
$\dfrac{1}{x}+\dfrac1y+\dfrac1z+\dfrac1t$
$=(\dfrac1x+\dfrac1y)+(\dfrac1z+\dfrac1t)$
$\ge \dfrac4{x+y}+\dfrac4{z+t}$
$\ge 4(\dfrac1{x+y}+\dfrac1{z+t})$
$\ge 4\cdot\dfrac4{x+y+z+t}$
$\ge \dfrac{16}{x+y+z+t}$
$\to (x+y+z+t)(\dfrac{1}{x}+\dfrac1y+\dfrac1z+\dfrac1t)\ge 16$
Dấu = xảy ra khi $x=y=z=t$