Cách gần tương tự:
\(A=\dfrac{x}{y+1}+\dfrac{y}{x+1}\)
\(A+2=\dfrac{x}{y+1}+1+\dfrac{y}{x+1}+1\)
\(A+2=\dfrac{x+y+1}{y+1}+\dfrac{x+y+1}{x+1}\)
\(A+2=\left(x+y+1\right)\left(\dfrac{1}{y+1}+\dfrac{1}{x+1}\right)\)
\(A+2\ge2\cdot\dfrac{4}{x+y+2}=2\cdot\dfrac{4}{3}=\dfrac{8}{3}\)
\(\Rightarrow A\ge\dfrac{2}{3}\Leftrightarrow x=y=\dfrac{1}{2}\)
p/s:Cần cm bđt \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)(khá dễ)
Áp dụng bất đẳng thức Cauchy-Schwarz ta có: \(NL=\dfrac{x}{y+1}+\dfrac{y}{x+1}=\dfrac{x^2}{xy+x}+\dfrac{y^2}{xy+y}\ge\dfrac{\left(x+y\right)^2}{2xy+x+y}=\dfrac{1}{2xy+1}\) Mặt khác theo AM-GM: \(xy\le\dfrac{\left(x+y\right)^2}{4}\Leftrightarrow2xy\le\dfrac{\left(x+y\right)^2}{2}\Leftrightarrow2xy\le\dfrac{1}{2}\) hay \(NL\ge\dfrac{1}{2xy+1}\ge\dfrac{1}{\dfrac{1}{2}+1}=\dfrac{1}{\dfrac{3}{2}}=\dfrac{2}{3}\).Dấu "=" xảy ra khi: \(x=y=\dfrac{1}{2}\)
