Question

cho x y là 2 số không âm thỏa mãn x+y=1. chứng minh:\(\dfrac{x}{y+1}+\dfrac{y}{x+1}\) \(\le\) 1

Answer 1

Do \(\left\{{}\begin{matrix}x;y\ge0\\x+y=1\end{matrix}\right.\) \(\Rightarrow0\le x;y\le1\) \(\Rightarrow\left\{{}\begin{matrix}x^2\le x\\y^2\le y\end{matrix}\right.\) 

\(\Rightarrow x^2+y^2\le x+y=1\)

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

\(=\dfrac{x^2+y^2+1}{xy+2}\le\dfrac{x^2+y^2+1}{2}\le\dfrac{1+1}{2}=1\) (đpcm)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(1;0\right);\left(0;1\right)\)