Question

[hữu phú]

chứng minh ggawngr thức sau\(\frac{1}{1-x^2}+\frac{1}{1-y^2}\ge\frac{2}{1-xy}\)

Answer 1

[Phạm Thành Đông]

Giả sử.\(\frac{1}{1-x^2}+\frac{1}{1-y^2}\ge\frac{2}{1-xy}\left(x\ge1;y\ge1\right)\)

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

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

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

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

\(\Leftrightarrow\left(2-x^2-y^2\right)\left(1-xy\right)\ge2\left(1-x^2+x^2y^2-y^2\right)\)

\(\Leftrightarrow2-2xy-x^2+x^3y-y^2+xy^3\)\(\ge2-2x^2+2x^2y^2-2y^2\)

\(\Leftrightarrow2-2xy-x^2+x^3y-y^2+xy^3\)\(-2+2x^2+2y^2-2x^2y^2\ge0\)

\(\Leftrightarrow x^2+y^2-2xy+xy\left(x^2+y^2\right)-2xy.xy\ge0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+xy\left(x^2-2xy+y^2\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2+xy\left(x-y\right)^2\ge0\)

\(\Leftrightarrow\left(xy+1\right)\left(x-y\right)^2\ge0\)

Ta có:

\(\left(x-y\right)^2\ge0\forall x;y\);\(xy\ge1\)(vì \(x\ge1;y\ge1\))\(\Rightarrow xy+1\ge2\forall x\ge1;y\ge1\)

Do đó: \(\left(xy+1\right)\left(x-y\right)^2\ge0\forall x\ge1;y\ge1\)(luôn đúng).

Dấu bằng xảy ra \(\Leftrightarrow x=y=1\).

Vậy \(\frac{1}{1-x^2}+\frac{1}{1-y^2}\ge\frac{2}{1-xy}\)với \(x\ge1;y\ge1\).