leanhthu

mình thử rồi nhưng k ra

a,b,c> 0

Chứng minh rằng: $\frac{\left ( b+c-a \right )^{2}}{\left ( b+c \right )^{2}+a^{2}}+\frac{\left ( c+a-b \right )^2}{\left ( c+a \right )^{2}+b^{2}}+\frac{\left (a+b-c \right )^{2}}{\left ( a+b \right )^{2}+c^{2}}\geq \frac{3}{5}$

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