Question

[Babys Cutes]

Đề bài : Cho a + b > 2 . Chứng minh : a^4 + b^4 > 2 .

Answer 1

[Phạm Nguyễn Tất Đạt]

Ta có:\(\left(a^2-b^2\right)^2\ge0\)

\(\Leftrightarrow a^4+b^4-2a^2b^2\ge0\)

\(\Leftrightarrow a^4+b^4\ge2a^2b^2\)

\(\Leftrightarrow a^4+b^4+a^4+b^4\ge a^4+2a^2b^2+b^4\)

\(\Leftrightarrow2\left(a^4+b^4\right)\ge\left(a^2+b^2\right)^2\)

\(\Leftrightarrow a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\)

CMTT\(\Rightarrow a^2+b^2>\dfrac{\left(a+b\right)^2}{2}=\dfrac{2^2}{2}=2\)

\(\Rightarrow a^4+b^4>\dfrac{2^2}{2}=2\left(đpcm\right)\)

Answer 2

[Hồng Quang]

≥ 2 chứ bạn