Question

[Lê Phúc Thuận]

Cho a,b la số nguyên dương.   a+b=1

Tìm GTLN : Q=a/(1+2a) +b/(1+2b)

Answer 1

[Đinh Đức Hùng]

Cách lầy nèk

\(Q=\frac{a}{1+2a}+\frac{b}{1+2b}\le\frac{a}{2\sqrt{2a}}+\frac{b}{2\sqrt{2b}}=\frac{\sqrt{a}}{2\sqrt{2}}+\frac{\sqrt{b}}{2\sqrt{2}}\)

\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{2}}=\frac{\sqrt{\frac{a}{2}}+\sqrt{\frac{b}{2}}}{2\sqrt{2}.\frac{1}{\sqrt{2}}}\le\frac{\frac{a+\frac{1}{2}}{2}+\frac{b+\frac{1}{2}}{2}}{2}=\frac{a+b+1}{4}=\frac{2}{4}=\frac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=\frac{1}{2}\)

Answer 2

[Lê Anh Tú]

Có a+b =1 

Áp dụng bất đẳng thức cô-si 

=> ab= \(\frac{\left(a+b\right)}{2}\) 

<=> ab= \(\frac{1}{2}\)

P=\(\frac{\left(ab+a+ab+b\right)}{\left(ab+a+b+1\right)}\)

= \(\frac{\left(2ab+1\right)}{\left(ab+2\right)}\)

=\(\frac{\left[\left(2ab+4\right)-3\right]}{\left(ab+2\right)}\)

=\(2+\left[\frac{-3}{\left(ab+2\right)}\right]\)
Có ab = \(\frac{1}{2}\)

\(ab+2\Leftarrow\frac{5}{2}\)

\(\frac{1}{\left(ab+2\right)}\ge\frac{2}{5}\)

\(\frac{-1}{\left(ab+2\right)}\Leftarrow\frac{-2}{5}\)

\(\frac{-3}{ \left(ab+2\right)}\Leftarrow\frac{-6}{5}\)

=> GTLN = \(\frac{-6}{5}+2=\frac{4}{5}\) tại \(a=b=\frac{1}{2}\)