bunhia:
\(\left(1+1+1\right)\left(a^4+b^4+c^4\right)\ge\left(a^2+b^2+c^2\right)^2\)
\(\Rightarrow a^4+b^4+c^4\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3}\)
Ta cm bđt sau:\(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow\dfrac{1}{2}\left(a-b\right)^2+\dfrac{1}{2}\left(b-c\right)^2+\dfrac{1}{2}\left(c-a\right)^2\ge0\)(tự khai triển luôn đúng)
\(\Rightarrow a^2+b^2+c^2\ge12\Rightarrow\dfrac{\left(a^2+b^2+c^2\right)^2}{3}\ge\dfrac{12^2}{3}=48\)
\(\Rightarrow a^4+b^4+c^4\ge48\)
Dấu ''='' xảy ra khi a=b=c=2
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