Question

Chứng minh: a⁴ +b⁴ +c² >= abc(a+b+c)

Answer 1

Sửa đề: Chứng minh \(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)

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Áp dụng BĐT Cauchy Shwarz, ta có:

\(\left(1+1+1\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2\ge\dfrac{\left(a+b+c\right)^4}{9}\)

Áp dụng BĐT Cauchy Shwarz và BĐT AM - GM, ta có:

(+) \(\left(1+1+1\right)\left(a^4+b^4+c^4\right)\ge\left(a^2+b^2+c^2\right)^2\)

\(\Leftrightarrow a^4+b^4+c^4\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3}\)

\(\ge\dfrac{\dfrac{\left(a+b+c\right)^4}{9}}{3}=\dfrac{\left(a+b+c\right)^3}{27}\times\left(a+b+c\right)\)

\(\ge\dfrac{\left(3\sqrt[3]{abc}\right)^3}{27}\times\left(a+b+c\right)=abc\left(a+b+c\right)\)

Dấu "=" xảy ra khi a = b = c