Question

Biết \(a+b=1\). Chứng minh rằng:

\(a,a^2+b^2\ge\dfrac{1}{2}\)

\(b,a^4+b^4\ge\dfrac{1}{8}\)

\(c,a^8+b^8\ge\dfrac{1}{128}\)

Answer 1

Áp dụng bđt Cauchy-Schwarz:

\(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}=\dfrac{1}{2}\)

\(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left(\dfrac{1}{2}\right)^2}{2}=\dfrac{1}{8}\)

\(a^8+b^8\ge\dfrac{\left(a^4+b^4\right)^2}{2}\ge\dfrac{\left(\dfrac{1}{8}\right)^2}{2}=\dfrac{1}{128}\)