ta có $2x^2+3xy+4y^2=2((x+\frac34y)^2+(\frac{\sqrt{23y}}4)^2)$tương tự, ta được $P=\sqrt2(\sqrt{(x+\frac34y)^2+(\frac{\sqrt{23y}}4)^2}+\sqrt{(x+\frac34y)^2+(\frac{\sqrt{23y}}4)^2}+\sqrt{(x+\frac34y)^2+(\frac{\sqrt{23y}}4)^2})$Áp dụng BĐT Mincopxki => $P\geq \sqrt2(\sqrt{(\frac74(x+y+z))^2+(\frac{\sqrt{23}}2(x+y+z))^2}$=> $P\geq\sqrt2.\sqrt\frac{72}{16}.(x+y+z)=3(x+y+z)$mà $x+y+z\geq \sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1$=>$P\geq3$ Dấu = xảy ra tại $x=y=z=\frac13$
ta có $2x^2+3xy+4y^2=2((x+\frac34y)^2+(\frac{\sqrt{23}
y}4)^2)$tương tự, ta được $P=\sqrt2(\sqrt{(x+\frac34y)^2+(\frac{\sqrt{23}
y}4)^2}+\sqrt{(
y+\frac34
z)^2+(\frac{\sqrt{23}
z}4)^2}+\sqrt{(
z+\frac34
x)^2+(\frac{\sqrt{23}
x}4)^2})$Áp dụng BĐT Mincopxki => $P\geq \sqrt2(\sqrt{(\frac74(x+y+z))^2+(\frac{\sqrt{23}}2(x+y+z))^2}$=> $P\geq\sqrt2.\sqrt\frac{72}{16}.(x+y+z)=3(x+y+z)$mà $x+y+z\geq \sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1$=>$P\geq3$ Dấu = xảy ra tại $x=y=z=\frac13$