Question

Cho x,y,z>0 thỏa mãn \(x^2+y^2+z^2+2xy=3\left(x+y+z\right)\).Tìm GTNN \(P=x+y+z+\frac{20}{\sqrt{x+z}}+\frac{20}{\sqrt{y+2}}\)

Answer 1

Số hạng cuối là \(\frac{20}{\sqrt{y+2}}\) hay \(\frac{20}{\sqrt{y+z}}\) vậy bạn?

Answer 2

\(3\left(x+y+z\right)=\left(x+y\right)^2+z^2\ge\frac{1}{2}\left(x+y+z\right)^2\)

\(\Rightarrow x+y+z\le6\)

\(P\ge x+y+z+\frac{80}{\sqrt{x+z}+\sqrt{y+2}}=x+y+z+\frac{320}{2.2\sqrt{x+z}+2.2\sqrt{y+2}}\)

\(P\ge x+y+z+\frac{320}{4+x+z+4+y+2}=x+y+z+\frac{320}{x+y+z+10}\)

\(P\ge x+y+z+10+\frac{256}{x+y+z+10}+\frac{64}{x+y+z+10}-10\)

\(P\ge2\sqrt{\frac{256\left(x+y+z+10\right)}{x+y+z+10}}+\frac{64}{6+10}-10=26\)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;2;3\right)\)