Cho a, b, c là các số thực dương thỏa mãn a + b + c = 6. Chứng minh

\(\frac{a}{{\sqrt {{b^3} + 1} }} + \frac{b}{{\sqrt {{c^3} + 1} }} + \frac{c}{{\sqrt {{a^3} + 1} }}\; \ge 2\)

Áp dụng BĐT Cô-si ta có:

\(\sqrt {{b^3} + 1} \) = \(\sqrt {\left( {b + 1} \right)\left( {{b^2} - b + 1} \right)} \) \( \le \) \(\frac{{b + 1 + \;{b^2} - b + 1}}{2} = \) \(\frac{{{b^2} + \;2}}{2}\)

T2: \(\sqrt {{c^3} + 1} \) \( \le \) \(\frac{{{c^2} + \;2}}{2}\)  ;  \(\sqrt {{a^3} + 1} \) \( \le \) \(\frac{{{a^2} + \;2}}{2}\) 

Do đó VT \( \ge \) \(\frac{{2a}}{{{b^2} + \;2}}\) + \(\frac{{2b}}{{{c^2} + \;2}}\) + \(\frac{{2c}}{{{a^2} + \;2}}\)

Ta cần CM: S = \(\frac{{2a}}{{{b^2} + \;2}}\) + \(\frac{{2b}}{{{c^2} + \;2}}\) + \(\frac{{2c}}{{{a^2} + \;2}}\) \( \ge \) 2

Ta có: \(\frac{{2a}}{{{b^2} + \;2}}\) = \(\frac{{a\left( {{b^2} + 2} \right) - a{b^2}}}{{{b^2} + \;2}}\) = a - \(\frac{{a{b^2}}}{{{b^2} + \;2}}\)   

Lại có : \(\frac{{a{b^2}}}{{{b^2} + \;2}}\) = \(\frac{{2a{b^2}}}{{{b^2} + {b^2} + \;4}}\) \( \le \) \(\frac{{2a{b^2}}}{{{3^3}\sqrt {{b^4}.\;\;4} }}\) = \(\frac{{{a^3}\sqrt {2{b^2}} }}{3}\)  \( \le \) \(\frac{{a.\left( {2 + b + b} \right)}}{9}\) = \(\frac{{2a.\left( {b + 1} \right)}}{9}\)

T2 ta được S \( \ge \) a + b + c - \(\frac{{2.\left( {a + b + c} \right)}}{9}\) - \(\frac{{2\left( {ab + bc + ca} \right)}}{9}\) = \(\frac{{7.\left( {a + b + c} \right)}}{9}\) - \(\frac{{2\left( {ab + bc + ca} \right)}}{9}\)

Ta có ab + bc + ca \( \le \) \(\frac{{{{\left( {a + b + c} \right)}^2}}}{3}\)

Do đó S \( \ge \) \(\frac{{7\;.\;6}}{9}\) - \(\frac{2}{9}\) . \(\frac{{{6^2}}}{9}\) = 2

Dấu bằng xảy ra khi a = b = c = 2. Ta có đpcm.