Question

Tìm GTNN của \(S=\sqrt{x-2}+\sqrt{y-3}\)

biết x+y=6 

Answer 1

\(S=\sqrt{x-2}+\sqrt{y-3}\)

\(\Rightarrow S^2=\left(\sqrt{x-2}+\sqrt{y-3}\right)^2\)

\(\Rightarrow S^2=x-2+2\sqrt{\left(x-2\right)\left(y-3\right)}+y-3\)

\(\Rightarrow S^2=x+y-5+2\sqrt{\left(x-2\right)\left(y-3\right)}\)

\(\Rightarrow S^2=1+2\sqrt{\left(x-2\right)\left(y-3\right)}\)

Vì \(2\sqrt{\left(x-2\right)\left(y-3\right)}\ge0\)

\(\Rightarrow1+2\sqrt{\left(x-2\right)\left(y-3\right)}\ge1\)

\(\Rightarrow S^2\ge1\Leftrightarrow\orbr{\begin{cases}S\ge1\left(tm\right)\\S\le-1\left(ktm\right)\end{cases}}\)

\(\Rightarrow S_{min}=1\Leftrightarrow2\sqrt{\left(x-2\right)\left(y-3\right)}=0\)

TH1 : \(x-2=0\Leftrightarrow x=2\Rightarrow y=6-2=4\)

Th2 : \(y-3=0\Rightarrow y=3\Rightarrow x=6-3=3\)

Vậy \(S_{min}=1\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}\)hoặc \(x=y=3\)

Answer 2

Áp dụng bđt Bu-nhi-a-cốp-xki ta có

\(S^2=\left(\sqrt{x-2}+\sqrt{y-3}\right)^2\le\left(1+1\right)\left(x+y-5\right)=2\left(6-5\right)=2\)(vì \(x+y=6\) )

\(\Rightarrow S^2\le2\)

\(\Leftrightarrow-\sqrt{2}\le S\le\sqrt{2}\)

\(\Rightarrow minS=-\sqrt{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{\sqrt{x-2}}{1}=\frac{\sqrt{y-3}}{1}\\x+y=6\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=2,5\\y=3,5\end{cases}}\)