Question

Cho 2x-y=2. TÌm Giá trị nhỏ nhất

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

Answer 1

Giải:

\(A=\sqrt{x^2+(y+1)^2}+\sqrt{x^2+(y-3)^2}\)

\(\Leftrightarrow A=\sqrt{x^2+(2x-1)^2}+\sqrt{x^2+(2x-5)^2}\)

ÁP dụng BĐT Cauchy-Schwarz:

\([x^2+(2x-1)^2](2^2+1)\geq (2x+2x-1)^2\Rightarrow \sqrt{x^2+(2x-1)^2}\geq \frac{|4x-1|}{\sqrt{5}}\)

\([x^2+(2x-5)^2](2^2+11^2)\geq (2x+55-22x)^2\Rightarrow \sqrt{x^2+(2x-5)^2}\geq \frac{|-20x+55|}{5\sqrt{5}}=\frac{|-4x+11|}{\sqrt{5}}\)

\(\Rightarrow A\geq \frac{|4x-1|+|-4x+11|}{\sqrt{5}}\geq \frac{|4x-1-4x+11|}{\sqrt{5}}=\frac{10}{\sqrt{5}}=2\sqrt{5}\)

Vậy \(A_{\min}=2\sqrt{5}\Leftrightarrow x=\frac{2}{3}\)