Cho $a,b,c$ là các số thực dương. Chứng minh rằng:
$\dfrac{a}{a^2+bc+b^2}+\dfrac{b}{b^2+ca+c^2}+ \dfrac{c}{c^2+ab+a^2} \le \dfrac{a+b+c}{ab+bc+ca}$
Không hiểu, mod sửa giúp!
hic hic mình học *** chỉ tới đây thôi tiếp ko đc nữa, mong cao nhân chỉ giám
\[\begin{array}{l}
\frac{a}{{{a^2} + bc + {b^2}}} + \frac{b}{{{b^2} + ca + {c^2}}} + \frac{c}{{{c^2} + ab + {a^2}}} \le \frac{{a + b + c}}{{ab + bc + ca}}\\
\left( {{a^2} + bc + {b^2}} \right)\left( {{c^2} + bc + {a^2}} \right) \ge {\left( {ac + bc + ba} \right)^2}\\
\left( {{b^2} + ca + {c^2}} \right)\left( {{a^2} + ca + {b^2}} \right) \ge {\left( {ba + ca + bc} \right)^2}\\
\left( {{c^2} + ab + {a^2}} \right)\left( {{b^2} + ab + {c^2}} \right) \ge {\left( {bc + ab + ca} \right)^2}\\
VT = \frac{{a\left( {{c^2} + bc + {a^2}} \right)}}{{\left( {{a^2} + bc + {b^2}} \right)\left( {{c^2} + bc + {a^2}} \right)}} + \frac{{b\left( {{a^2} + ca + {b^2}} \right)}}{{\left( {{b^2} + ca + {c^2}} \right)\left( {{a^2} + ca + {b^2}} \right)}} + \frac{{c\left( {{b^2} + ab + {c^2}} \right)}}{{\left( {{c^2} + ab + {a^2}} \right)\left( {{b^2} + ab + {c^2}} \right)}}\\
\le \frac{{a\left( {{c^2} + bc + {a^2}} \right) + b\left( {{a^2} + ca + {b^2}} \right) + c\left( {{b^2} + ab + {c^2}} \right)}}{{{{\left( {ab + bc + ca} \right)}^2}}}\\
{c^2} + bc + {a^2} \le \sqrt {\left( {{c^2} + {b^2} + {a^2}} \right)\left( {2{c^2} + {a^2}} \right)} \\
{a^2} + ca + {b^2} \le \sqrt {\left( {{a^2} + {c^2} + {b^2}} \right)\left( {2{a^2} + {b^2}} \right)} \\
{b^2} + ab + {c^2} \le \sqrt {\left( {{b^2} + {a^2} + {c^2}} \right)\left( {2{b^2} + {c^2}} \right)} \\
VT \le \frac{{\sqrt {{a^2} + {b^2} + {c^2}} \left( {a\sqrt {2{c^2} + {a^2}} + b\sqrt {2{a^2} + {b^2}} + c\sqrt {2{b^2} + {c^2}} } \right)}}{{{{\left( {ab + bc + ca} \right)}^2}}}\\
\le \frac{{2\left( {{a^2} + {b^2} + {c^2}} \right)\sqrt {{a^2} + {b^2} + {c^2}} }}{{{{\left( {ab + bc + ca} \right)}^2}}}
\end{array}\]