Question

[phùng thị khánh huyền]

cho x.y>0 và x+y=1

cm\((x+\frac{1}{x})^2+(y+\frac{1}{y})^2\ge\frac{25}{2}\)

Answer 1

[l҉o҉n҉g҉ d҉z҉]

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

\(\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\ge\frac{\left(x+\frac{1}{x}+y+\frac{1}{y}\right)^2}{2}\ge\frac{\left(1+\frac{4}{x+y}\right)^2}{2}=\frac{\left(1+4\right)^2}{2}=\frac{25}{2}\)

=> đpcm . Dấu "=" xảy ra <=> x = y = 1/2

Answer 2

[ミ★Zero ❄ ( Hoàng Nhật )]

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

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

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

Áp dụng BĐT Co-si , có :

\(x^2+y^2\ge2xy\)

\(\Rightarrow2\left(x^2+y^2\right)\ge x^2+y^2+2xy\)

\(\Leftrightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)

\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)

Có \(\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{2}{xy}\ge\frac{8}{\left(x+y\right)^2}\)

\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}\ge8\)

\(\Rightarrow A\ge\frac{1}{2}+8+4\)

\(\Rightarrow A\ge\frac{25}{2}\)