Question

Cho x , y , z > 0 thỏa mãn : x + 2y + 3z = 3 

Tìm min \(\frac{x}{1+4y^2}+\frac{2y}{1+9z^2}+\frac{3z}{1+x^2}\)

 

Answer 1

Đặt \(x=a;2y=b;3z=c\Rightarrow a+b+c=3\) 

\(T=\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\)

Áp dụng Bđt Cô si ngược dấu ta có:

\(T=\text{∑}a-\frac{a^2b}{1+b^2}\ge\text{∑}a-\frac{a^2b}{2b}=\text{∑}a-\frac{ab}{2}\)

\(=a+b+c-\frac{ab+bc+ca}{2}\ge a+b+c-\frac{\left(ab+bc+ca\right)^2}{6}\)\(=3-\frac{3^2}{6}=\frac{3}{2}\)

Dấu = khi \(a=b=c=1\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{2}\\z=\frac{1}{3}\end{cases}}\)