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*Áp dụng Cosi với x,y>0 ta có:
\(x+y\ge2\sqrt{xy}\left(1\right)\)
\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\left(2\right)\)
Nhân (1),(2) có: \(\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\RightarrowĐPCM\)
**\(\frac{1}{xy}+\frac{1}{x\left(x+y\right)}+\frac{1}{y\left(x+y\right)}+\frac{1}{x^2+y^2}\)
Ta có: \(\frac{1}{x\left(x+y\right)}+\frac{1}{y\left(x+y\right)}\ge\frac{4}{x^2+2xy+y^2}=4\)
Có: \(\frac{1}{x^2+xy}+\frac{1}{y^2+xy}\ge\frac{4}{\left(x+y\right)^2}\le4\)
Theo Cosi ta có: \(xy\le\left(\frac{x+y}{2}\right)^2\)
\(\Rightarrow\frac{1}{xy}\ge\left(\frac{2}{x+y}\right)^2\ge\left(\frac{2}{1}\right)^2=4\)
Áp dụng Cosi ta có: \(2xy\left(x^2+y^2\right)\le\left(\frac{x^2+2xy+y^2}{2}\right)^2=\frac{\left(x+y\right)^4}{4}\le\frac{1}{4}\)
\(\Rightarrow xy\left(x^2+y^2\right)\le\frac{1}{8}\)(1)
Mà ta có ở trên: \(xy\le\frac{\left(x+y\right)^2}{4}\le\frac{1}{4}\)(2)
Từ (1) và (2) ta có: \(x^2+y^2\le\frac{1}{2}\Rightarrow\frac{1}{x^2+y^2}\ge2\)
Vậy Ta có: \(\frac{1}{xy}+\frac{1}{x^2+xy}+\frac{1}{y^2+xy}+\frac{1}{x^2+y^2}\ge4+4+2=10\)
Với x=y=1/2
quy đồng BĐT \(\frac{\left(xy-1\right)\left(x-y\right)^2}{\left(x^2+1\right)\left(y^2+1\right)\left(xy+1\right)}\ge0\forall xy\ge1\)
Ta có: \(\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge\frac{2}{1+xy}\)
\(\Leftrightarrow\left(\frac{1}{1+x^2}-\frac{1}{1+xy}\right)+\left(\frac{1}{1+y^2}-\frac{1}{xy}\right)\ge0\)
\(\Leftrightarrow\frac{xy-x^2}{\left(1+x^2\right)\left(1+xy\right)}+\frac{xy-y^2}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow x\left(y-x\right)\left(1+y^2\right)+y\left(x-y\right)\left(1+x^2\right)\ge0\)
\(\Leftrightarrow\left(y-x\right)\left(x+xy^2-y-x^2y\right)\ge0\)
\(\Leftrightarrow\left(y-x\right)^2\left(xy-1\right)\ge0\)(đúng với mọi x,y>=1)
Chuyển vế biến đổi tương đương
\(\frac{1}{1+x^2}-\frac{1}{xy+1}+\frac{1}{1+y^2}-\frac{1}{xy+1}\ge0\)
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Áp dụng BĐT : ( a + b + c )2 \(\ge\)3 ( ab + bc + ac )
Ta có : \(\frac{\left(x+y+1\right)^2}{xy+y+x}\ge\frac{3\left(xy+y+x\right)}{xy+y+x}=3\)
đặt \(\frac{\left(x+y+1\right)^2}{xy+y+x}=A\)
ta có : \(A+\frac{1}{A}=\frac{8A}{9}+\frac{A}{9}+\frac{1}{A}\ge\frac{8.3}{9}+2\sqrt{\frac{A}{9}.\frac{1}{A}}=\frac{8}{3}+\frac{2}{3}=\frac{10}{3}\)
Ta có \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
=> \(a^2+b^2+c^2\ge ab+bc+ac\)=> \(\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)
Áp dụng ta được
\(\left(x+y+1\right)^2\ge3\left(x+y+xy\right)\)=> \(\frac{\left(x+y+1\right)^2}{xy+y+x}\ge3\)
Đặt \(\frac{\left(x+y+1\right)^2}{x+y+xy}=t\)(\(t\ge3\))
Khi đó
\(VT=t+\frac{1}{t}=\left(\frac{t}{9}+\frac{1}{t}\right)+\frac{8}{9}t\ge\frac{2}{3}+\frac{8}{9}.3=\frac{10}{3}\)
Dấu bằng xảy ra khi \(\hept{\begin{cases}t=3\\x=y=1\end{cases}}\)=> x=y=1
Lưu ý
Nhiều người sẽ nhầm \(VT\ge2\)
Khi đó dấu bằng \(\left(x+y+1\right)^2=xy+x+y\)không xảy ra
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 \(x\ge1;y\ge1\).