\(\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}+\sqrt{\left(y+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}_{ }+\sqrt{\left(z-2\right)^2+\left(\sqrt{3}\right)^2}\ge.\)

\(\sqrt{\left(x+y+1\right)^2+\left(\sqrt{3}\right)^2}+\sqrt{\left(z-2\right)^2+\left(\sqrt{3}\right)^2}\ge\sqrt{\left(x+y+z-1\right)^2+12}=4.\)
Sử dụng Minkowski,

(căn x+căn y+căn z)^2=1

=> x+y+z+2(căn xy+căn yz+căn xz)=1

mà căn xy+căn yz+ căn xz =< x+y+z ( nhân cả 2 vế với 2 để c/m)

=>3(x+y+z)>=1

=>x+y+z>=1/3

dấu = xảy ra khi x=y=z=1/9

x=y=z=1/9

=>x+y+z=1/3

a: ĐKXĐ: x>=0; x<>1

\(P=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\)

\(=\frac{3+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\left(\sqrt{x}+1\right)\)

\(=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)

b: Thay \(x=3+2\sqrt2=\left(\sqrt2+1\right)^2\) vào P, ta được:

\(P=\frac{\sqrt{\left(\sqrt2+1\right)^2}+2}{\sqrt{\left(\sqrt2+1\right)^2}-1}=\frac{\sqrt2+1+2}{\sqrt2+1-1}=\frac{3+\sqrt2}{\sqrt2}=\frac{3\sqrt2+2}{2}\)

c: P<0

=>\(\frac{\sqrt{x}+2}{\sqrt{x}-1}<0\)

=>\(\sqrt{x}-1<0\)

=>\(\sqrt{x}<1\)

=>0<=x<1