bài 3:
a, ta cần c/m $\left ( a+b+c-2abc \right )^{2}\leq 2$
ta có 1=$a^{2}+b^{2}+c^{2}\geq a^{2}+2bc\geq 2bc\Rightarrow 1-2bc\geq 0$$a^{2}+b^{2}+c^{2}\geq a^{2}+2bc\geq 2bc\Rightarrow 1-2bc\geq 0$ (*).Dễ thấy $\left ( a+b+c-2abc \right )^{2}=\left [ a\left ( 1-2bc \right )+\left ( b+c \right ) \right ]^{2}$.Áp dụng bđt bunhia ta có
$(a+b+c-2abc)^{2}\leq \left [ a^{2}+\left ( b+c \right )^{2}\right]\left [ \left ( 1-2bc \right )^{2} +1\right ]\left ( 1 \right )$
vì $a^{2}+b^{2}+c^{2}=1$ nên $\left ( 1 \right ) \Leftrightarrow \left ( a+b+c-2abc \right )^{2} \leq \left ( 1+2bc \right )\left ( 2-4bc+4b^{2}c^{2} \right )\left ( 2 \right )$. Do $\left ( 1+2bc \right )\left ( 2-4bc+4b^{2} c^{2}\right )=2-4b^{2}c^{2}\left ( 1-2bc \right )$$\left ( 1+2bc \right )\left ( 2-4bc+4b^{2} c^{2}\right )=2-4b^{2}c^{2}\left ( 1-2bc \right )$. Từ (*)$\Rightarrow 4b^{2}c^{2}\left ( 1-2bc \right )\geq 0$.Do đó $\left ( 1+2bc \right )\left ( 2-4bc-4b^{2}c^{2} \right )\leq 2$$\left ( 1+2bc \right )\left ( 2-4bc-4b^{2}c^{2} \right )\leq 2$. Từ (2) ta có $\left ( a+b+c-2abc \right )^{2}\leq 2 \Leftrightarrow a+b+c\leq 2abc +\sqrt{2}$(dpcm)
Bài viết đã được chỉnh sửa nội dung bởi ductai202: 06-08-2014 - 20:55