leanhthu

Cho a,b,c là 3 cạnh của 1 tam giác. Chứng minh:

$\frac{a^{2016}}{b+c-a}+\frac{b^{2016}}{c+a-b}+\frac{c^{2016}}{a+b-c}\geq a^{2016}+b^{2016}+c^{2016}$

BĐT $\Leftrightarrow \frac{1}{bc\cdot \left ( 3a+b+c \right )}+\frac{1}{ac\cdot \left ( a+3b+c \right )}+\frac{1}{ab\left ( a+b+3c \right )}\geq \frac{24}{5(a+b)\cdot \left ( b+c \right )\cdot \left ( c+a \right )}$

Áp dụng bất đẳng thức cauchy - schwartz, ta có:

$ \frac{1}{bc\cdot \left ( 3a+b+c \right )}+\frac{1}{ac\cdot \left ( a+3b+c \right )}+\frac{1}{ab\left ( a+b+3c \right )}\geqslant \frac{9}{(a+b)(b+c)(c+a)+7abc}$

ta có : $ \frac{9}{(a+b)(b+c)(c+a)+7abc}$$ \geq \frac{24}{5(a+b)(b+c)(c+a)}$$ \Leftrightarrow (a+b)(b+c)(c+a)\geq 8abc$

Dấu bằng <=> a=b=c