5) Cho $\left\{\begin{matrix}a,b,c>0 & & \\ a+b+c\leq \frac{3}{2} & & \end{matrix}\right.$. Tìm Min $S=a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Ta phân tích thành : $(a+\frac{1}{a})+(b+\frac{1}{b})+(c+\frac{1}{c})=(4a+\frac{1}{a})-3a+(4b+\frac{1}{b})-3b+(4c+\frac{1}{c})-3c \geq 2.2+2.2+2.2-3(a+b+c)=12-\frac{9}{2}=\frac{15}{2}$
dấu bằng xảy ra khi $x=y=z=\frac{1}{2}$