Sory bài làm bị lỗi, gửi lại:
Áp dụng BĐT Cauchy - Schwarz dạng Engel:
\(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{1+1}=\dfrac{1}{2}\)
\(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{1+1}\ge\dfrac{\left(\dfrac{1}{2}\right)^2}{2}=\dfrac{1}{8}\)
\("="\Leftrightarrow a=b=\dfrac{1}{2}\)
Áp dụng BĐT Cauchy - Schwarz dạng Engel
\(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}=\dfrac{1}{2}\)
\(Cauchy-Schwarz:\)
\(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{1+1}\ge\dfrac{\left(\dfrac{1}{2}\right)^2}{1+1}=\dfrac{1}{8}\)
\("="\Leftrightarrow a=b=\dfrac{1}{2}\)
Áp dụng BĐT Cauchy - Schwarz dạng Engel:
\(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{1+1}=\dfrac{1}{2}\)
Lại áp dụng lần nữa, ta có:
\(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{1+1}=\dfrac{\left(\dfrac{1}{2}\right)^2}{2}=\dfrac{1}{8}\)
\("="\Leftrightarrow a=b=\dfrac{1}{2}\)
