Question

Cho 2 số dương x,y thỏa mãn x+y=1
tìm min A=(x+\(\frac{1}{x}\))²+(y+\(\frac{1}{y}\))²

Answer 1

Ta có:\(A=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

\(A=x^2+\frac{1}{x^2}+y^2+\frac{1}{y^2}+4\)

\(A=x^2+\frac{1}{16x^2}+y^2+\frac{1}{16y^2}+\frac{15}{16}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+4\)

\(A\ge\frac{1}{2}+\frac{1}{2}+\frac{15}{16}\cdot\frac{\left(\frac{1}{x}+\frac{1}{y}\right)^2}{2}+4\)

\(A\ge\frac{15}{32}\cdot\left(\frac{4}{x+y}\right)^2+5=\frac{15}{32}\cdot16+5=\frac{25}{2}\)

"="<=>x=y=1/2