Question

Cho a, b, c > 0 thỏa mãn ab+bc+ca=12

Tìm GTNN của M=a⁴+b⁴+c⁴

Answer 1

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