Question

cho 2 số nguyên a,b bất kì .Chứng minh: \(a^4+b^4\ge\frac{\left(a+b\right)^4}{8}\)

HELPPPPPPPPPP

Answer 1

Áp dụng BĐT Cauchy Schwarz và BĐT \(a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\), ta có:

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

\(\Leftrightarrow a^4+b^4\ge\dfrac{\left(a+b\right)^4}{8}\)

Dấu "=" xảy ra khi a = b

Answer 2

Áp dụng BĐT Bunhiacopxki,ta có:

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

Dấu = xảy ra khi a=b