Mình nghĩ đề đúng phải là: $P=2\left(x^2+y^2+z^2+t^2\right)+3\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right)\geq\dfrac{97}{2}$
Ta đặt: $s=x+y+z+t$.
Ta có:
$4(x^2+y^2+z^2+t^2)\geq(x+y+z+t)^2 \Leftrightarrow x^2+y^2+z^2+t^2\geq \dfrac{s^2}{4}$
$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\geq\dfrac{16}{x+y+z+t}=\dfrac{16}{s}$
Suy ra:
$A\geq\dfrac{s^2}{2}+\dfrac{48}{s}$
$=\dfrac{s^2}{2}+\dfrac{1}{2s}+\dfrac{1}{2s}+\dfrac{47}{s}$
$\ge3\sqrt[3]{\dfrac{s^2}{2}.\dfrac{1}{2s}.\dfrac{1}{2s}}+47=\dfrac{97}{2}$
Dấu bằng xảy ra khi: $x=y=z=t=\dfrac{1}{4}$