Áp dụng bđt \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) ta có :
\(\frac{a}{1+2a}\le\frac{1}{4}\left(\frac{a}{1}+\frac{a}{2a}\right)=\frac{1}{4}\left(a+\frac{1}{2}\right)\)
\(\frac{b}{1+2b}\le\frac{1}{4}\left(\frac{b}{1}+\frac{b}{2b}\right)=\frac{1}{4}\left(b+\frac{1}{2}\right)\)
Cộng vế với vế ta được :
\(\frac{a}{1+2a}+\frac{b}{1+2b}\le\frac{1}{4}\left(a+b+1\right)=\frac{1}{4}.2=\frac{1}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=\frac{1}{2}\)
\(\frac{a}{1}+\frac{a}{2b}\ge\frac{\left(2a\right)^2}{1+2b}=\frac{4a^2}{1+2b}.\)
hùng sai nhé phải là căn A nhé
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Rightarrow\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\ge\frac{1}{x+y}\) hay \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Sai cc
