Question

Cho các số thực dương a,b,c thỏa mãn: \(a^2+b^2+c^2+2ab=3\left(a+b+c\right)\). Tìm min của biểu thức: \(P=a+b+c+\frac{20}{\sqrt{a+c}}+\frac{20}{\sqrt{b+2}}\)

Answer 1

\(3\left(a+b+c\right)=\left(a+b\right)^2+c^2\ge\frac{1}{2}\left(a+b+c\right)^2\)

\(\Rightarrow a+b+c\le6\)

\(P=a+b+c+\frac{80}{2.2.\sqrt{a+c}}+\frac{80}{2.2\sqrt{b+2}}\ge a+b+c+\frac{80}{a+c+4}+\frac{80}{b+6}\)

\(P\ge a+b+c+\frac{320}{a+b+c+10}\)

\(P\ge a+b+c+10+\frac{256}{a+b+c+10}+\frac{64}{a+b+c+10}-10\)

\(P\ge2\sqrt{\frac{256\left(a+b+c+10\right)}{a+b+c+10}}+\frac{64}{6+10}-10=26\)

\(P_{min}=26\) khi \(\left(a;b;c\right)=\left(1;2;3\right)\)