Question

[Nguyễn Thiều Công Thành]

cho a;b;c là các số thực dương thỏa mãn abc=8.Tìm Max P=\(\frac{1}{2a+b+6}+\frac{1}{2b+c+6}+\frac{1}{2c+a+6}\)

Answer 1

[Thắng Nguyễn]

Áp dụng BĐT Cauchy-Schwarz ta có:

\(P=\frac{1}{\left(a+2\right)+\left(a+2\right)+\left(b+2\right)}+\frac{1}{\left(b+2\right)+\left(b+2\right)+\left(c+2\right)}+\frac{1}{\left(c+2\right)+\left(c+2\right)+\left(a+2\right)}\)

\(\le\frac{1}{9}\left(\frac{2}{a+2}+\frac{1}{b+2}\right)+\frac{1}{9}\left(\frac{2}{b+2}+\frac{1}{c+2}\right)+\frac{1}{9}\left(\frac{2}{c+2}+\frac{1}{a+2}\right)\)

\(=\frac{1}{3}\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)\)

Dễ dàng cm BĐT \(\frac{1}{x+1}+\frac{1}{y+1}\ge\frac{2}{1+\sqrt{xy}}\)

\(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=\frac{1}{2}\left(\frac{1}{1+\frac{a}{2}}+\frac{1}{1+\frac{b}{2}}+\frac{1}{1+\frac{c}{2}}\right)\)

\(\le\frac{1}{2}.\frac{3}{1+\sqrt[3]{\frac{abc}{8}}}=\frac{3}{4}\Rightarrow P\le\frac{1}{4}\)

Xảy ra khi \(a=b=c=2\)

Answer 2

[Thắng Nguyễn]

À viết ngược dấu BĐT phụ r` :v

\(\frac{1}{1+x}+\frac{1}{1+y}\le\frac{2}{1+\sqrt{xy}}\) mới đúng nhé :v

\(\Leftrightarrow\frac{\left(\sqrt{xy}-1\right)\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(x+1\right)\left(y+1\right)\left(1+\sqrt{xy}\right)}\le0\)