Với các số thực dương a, b, c thỏa mãn \(a+b+c=1\)1) Chứng minh rằng \(\frac{{{a^2}}}{b} + \frac{{{b^2}}}{c} + \frac{{{c^2}}}{a} \ge

1) Áp dụng bất đẳng thức AM – GM  ta có

\(\frac{{{a^2}}}{b} + b + \frac{{{b^2}}}{c} + c + \frac{{{c^2}}}{a} + a \ge 2\sqrt {\frac{{{a^2}}}{b}.b}  + 2\sqrt {\frac{{{b^2}}}{c}.c}  + 2\sqrt {\frac{{{c^2}}}{a}.a}  = 2a + 2b + 2c.\)

Do đó \(\frac{{{a^2}}}{b} + \frac{{{b^2}}}{c} + \frac{{{c^2}}}{a} \ge a + b + c = 1.\)

2) Ta chứng mình \(\frac{{{a^2}}}{b} + \frac{{{b^2}}}{c} + \frac{{{c^2}}}{a} \ge 3\left( {{a^2} + {b^2} + {c^2}} \right)\) với \(a + b + c = 1.\)

Thật vậy, bất đẳng thức tương tương

\(\begin{array}{l}
\left( {\frac{{{a^2}}}{b} - 2a + b} \right) + \left( {\frac{{{b^2}}}{c} - 2b + c} \right) + \left( {\frac{{{c^2}}}{a} - 2c + a} \right) \ge 3\left( {{a^2} + {b^2} + {c^2}} \right) - {\left( {a + b + c} \right)^2}\\
 \Leftrightarrow \frac{{{{\left( {a - b} \right)}^2}}}{b} + \frac{{{{\left( {b - c} \right)}^2}}}{c} + \frac{{{{\left( {c - a} \right)}^2}}}{a} \ge {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2}
\end{array}\)

Hay \(\left( {\frac{1}{b} - 1} \right){\left( {a - b} \right)^2} + \left( {\frac{1}{c} - 1} \right){\left( {b - c} \right)^2} + \left( {\frac{1}{a} - 1} \right){\left( {c - a} \right)^2} \ge 0\) luôn đúng do \(0 < a,b,c < 1.\)

Do đó \(P = 2017\left( {\frac{{{a^2}}}{b} + \frac{{{b^2}}}{c} + \frac{{{c^2}}}{a}} \right) + \frac{{{a^2}}}{b} + \frac{{{b^2}}}{c} + \frac{{{c^2}}}{a} + \frac{1}{{3\left( {{a^2} + {b^2} + {c^2}} \right)}}\)

\(\begin{array}{l}
 \ge 2017\left( {\frac{{{a^2}}}{b} + \frac{{{b^2}}}{c} + \frac{{{c^2}}}{a}} \right) + 3\left( {{a^2} + {b^2} + {c^2}} \right) + \frac{1}{{3\left( {{a^2} + {b^2} + {c^2}} \right)}}\\
\mathop  \ge \limits^{AM - GM} 2017 + 2.\sqrt {3\left( {{a^2} + {b^2} + {c^2}} \right) + \frac{1}{{3\left( {{a^2} + {b^2} + {c^2}} \right)}}}  = 2019.
\end{array}\)

Vậy \(\min P = 2019\) khi \(a = b = c = \frac{1}{3}.\)