Question

Cho x và y là các số thực không âm thỏa mãn x + y ≤ 2. Chứng minh: \(\frac{2+x}{1+x}+\frac{1-2y}{1+2y}\) ≥ \(\frac{8}{7}\)

Answer 1

\(\frac{2+x}{x+1}+\frac{1-2y}{1+2y}\)

\(=1+\frac{1}{x+1}-1+\frac{2}{1+2y}\)

\(=\frac{1}{x+1}+\frac{1}{\frac{1}{2}+y}\)

Áp dụng BDDT AM-GM ta có:

\(\frac{1}{x+1}+\frac{1}{\frac{1}{2}+y}\ge\frac{2}{\sqrt{\left(x+1\right).\left(\frac{1}{2}+y\right)}}\ge\frac{4}{x+1+\frac{1}{2}+y}\ge\frac{4}{\frac{3}{2}+2}=\frac{4}{\frac{7}{2}}=\frac{8}{7}\)

Dấu " = " xảy ra <=> \(\frac{1}{x+1}=\frac{1}{\frac{1}{2}+y}\); x+y=2

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+y=2\\y-x=\frac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\frac{5}{4}\\y=\frac{3}{4}\end{matrix}\right.\)