Question

Cho x, y là hai số thực dương sao cho x + y= 1

Chứng minh: \(\dfrac{x}{1-x^2}+\dfrac{y}{1-y^2}\ge\dfrac{4}{3}\)

Answer 1

ta có:

\(\dfrac{x}{1-x^2}+\dfrac{y}{1-y^2}=\dfrac{x-xy^2+y-x^2y}{\left(1-x^2\right)\left(1-y^2\right)}=\dfrac{1-xy}{xy\left(x+1\right)\left(y+1\right)}\)

Áp dụng BĐT cauchy:

\(\left(x+y\right)^2\ge4xy\Leftrightarrow xy\le\dfrac{1}{4}\)

và \(\left(x+1\right)\left(y+1\right)\le\dfrac{1}{4}\left(x+y+2\right)^2=\dfrac{9}{4}\)

do đó \(VT\ge\dfrac{1-\dfrac{1}{4}}{\dfrac{1}{4}.\dfrac{9}{4}}=\dfrac{3}{4}.\dfrac{16}{9}=\dfrac{4}{3}\)

dấu = xảy ra khi x=y=\(\dfrac{1}{2}\)