Đáp án: Giải thích các bước giải: 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\(\Rightarrow a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}=\dfrac{2^2}{2}=2\) \(\Rightarrow a^4+b^4\ge\dfrac{2^2}{2}=2\left(đpcm\right)\) Bình luận
Đáp án:
Giải thích các bước giải:
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\(\Rightarrow a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}=\dfrac{2^2}{2}=2\)
\(\Rightarrow a^4+b^4\ge\dfrac{2^2}{2}=2\left(đpcm\right)\)