Giải thích các bước giải:
Ta có :
$1=2\sqrt{xy}+\sqrt{xz}\le x+y+\dfrac{x+z}{2}=\dfrac{3x+2y+z}{2}$
$\to 3x+2y+z\ge 2$
Ta có :
$P=\dfrac{3yz}{x}+\dfrac{4zx}{y}+\dfrac{5xy}{z}$
$\to P=(\dfrac{yz}{x}+\dfrac{zx}{y})+2(\dfrac{yz}{x}+\dfrac{xy}{z})+3(\dfrac{xz}{y}+\dfrac{xy}{z})$
$\to P\ge 2\sqrt{\frac{yz}{x}.\frac{zx}{y}}+2.2\sqrt{\dfrac{yz}{x}.\dfrac{xy}{z}}+3.2\sqrt{\dfrac{xz}{y}.\dfrac{xy}{z}}= 2z+4y+6x=2(3x+2y+z)\geq 2.2=4$
Dấu = xảy ra khi $x=y=z=\dfrac 13$
