2/ Ta có : $\frac{1}{8x^{2}+1}-\frac{2}{x+1}+1\geq 0\Leftrightarrow \frac{2x\left ( 2x-1 \right )^{2}}{\left ( 8x^{2}+1 \right )\left ( x+1 \right )}\geq 0$
Tương tự với $b,c$ $\Rightarrow \sum \frac{1}{8x^{2}+1}\geq 2\sum \frac{1}{x+1}-3=1$
Đẳng thức xảy ra $\Leftrightarrow a=b=c=\frac{1}{2}$
3/ Ta có : $\frac{2a^{2}}{a+b^{2}}=2a-\frac{2ab^{2}}{a+b^{2}}\geq 2a-b\sqrt{a}\geq 2a-\frac{b+ab}{2}$
Tương tự $\Rightarrow \sum \frac{2a^{2}}{a+b^{2}}\geq \frac{3}{2}\sum a-\frac{1}{2}\sum ab\geq \frac{3}{2}\sum a -\frac{1}{6}\left ( a+b+c \right )^{2}=\frac{1}{2}\left ( 3-\sum a \right )\sum a +\sum a\geq a+b+c$
Do $a+b+c\geq 3$
Đẳng thức xay ra $\Leftrightarrow a=b=c=1$
5/ Ta có : $\frac{a^{2}}{\left ( a+b \right )^{2}}+\frac{b^{2}}{\left ( b+c \right )^{2}}=\frac{1}{\left ( 1+x^{2} \right )}+\frac{1}{\left ( 1+y \right )^{2}}\geq \frac{1}{1+xy}\Leftrightarrow \frac{\left ( x-y \right )^{2}}{\left ( 1+x \right )^{2}\left ( 1+y \right )^{2}}\geq 0$
với $x=\frac{b}{a},y=\frac{c}{b}\Rightarrow xy=\frac{c}{a}$
$\Rightarrow A\geq \frac{1}{1+xy}+\frac{xy}{4}=\frac{3}{4}+\frac{\left ( xy -1\right )^{2}}{4\left ( 1+xy \right )}\geq \frac{3}{4}$
Đẳng thức xay ra $\Leftrightarrow a=b=c$
6/ Ta có : $\frac{2}{\sqrt{\left ( 2x+y \right )^{3}+1}-1}+\frac{2}{\sqrt{\left ( 2y+x \right )^{3}+1}-1}\geq \frac{4}{\left ( 2x+y \right )^{2}}+\frac{4}{\left ( 2y+x \right )^{2}}$
Do $\sqrt{a^{3}+1}=\sqrt{\left ( a+1 \right )\left ( a^{2}-a+1 \right )}\leq \frac{a^{2}+2}{2}$
$\Rightarrow A\geq \frac{4}{\left ( 2x+y \right )^{2}}+\frac{4}{\left ( 2y+x \right )^{2}}+\frac{\left ( 2y+x \right )\left ( 2x+y \right )}{4}-\frac{8}{\left ( 2x+y \right )+\left ( 2y+x \right )}$
$\geq \frac{4}{(2x+y)(2y+x)}+\frac{16}{\left ( 2x+y+2y+x \right )^{2}}+\frac{(2x+y)(2y+x)}{4}-\frac{8}{2x+y+2y+x}\geq 1$
Đẳng thức xảy ra $\Leftrightarrow x=y=\frac{2}{3}$
Bài viết đã được chỉnh sửa nội dung bởi Senju Hashirama: 29-09-2016 - 13:18