Question

Cho \(a,b,c\) dương thỏa mãn \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=8\). Chứng minh

\(\dfrac{a+b+c}{3}\ge\sqrt[27]{\dfrac{a^3+b^3+c^3}{3}}\)

eztosol

Answer 1

\(\left(a^3+b^3+c^3\right)\left[\dfrac{3}{8}.\left(a+b\right)\left(b+c\right)\left(c+a\right)\right].\left[\dfrac{3}{8}\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]..\)

\(\le\dfrac{1}{9^9}\left[a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^9\)

\(=\dfrac{1}{9^9}\left(a+b+c\right)^{27}\)

\(\Leftrightarrow3^8.\left(a^3+b^3+c^3\right)\le\dfrac{1}{3^{18}}\left(a+b+c\right)^{27}\)

\(\Leftrightarrow\sqrt[27]{\dfrac{a^3+b^3+c^3}{3}}\le\dfrac{a+b+c}{3}\)

P/s: Eztogiveup