$a+b+c=3 CMR \sum \frac{a}{a^{3}+b^{2}+c}\leq 1$
theo cauchy-schwaz:
$\sum \frac{a}{a^{3}+b^{2}+c}$=$\sum \frac{a\left ( \frac{1}{a} \right +1+c)}{\left ( a^{3} \right+b^{2}+c )\left ( \frac{1}{a} +1+c\right )}\leq \frac{a\left ( \frac{1}{a} \right +1+c)}{\left ( a+b+c \right )^{2}}$=$\frac{1+a+ca}{9}$
ta chỉ cần chứng minh:
$\sum \frac{1+a+ca}{9}\leq 1 \Leftrightarrow ab+bc+ca\leq 3$
mà ta có:
$ab+bc+ca\leq \frac{\left ( a+b+c \right )^{2}}{3}$
từ đay suy ra ĐPCM "=" <=> a=b=c=1
$a+b+c=3 CMR \sum \frac{a}{a^{3}+b^{2}+c}\leq 1$
theo cauchy-schwaz:
\sum \frac{a}{a^{3}+b^{2}+c}\leq \frac{a\left ( \frac{1}{a} \right +1+c)}{\left ( a^{3} \right+b^{2}+c )\left ( \frac{1}{a} +1+c\right )}\leq \frac{a\left ( \frac{1}{a} \right +1+c)}{\left ( a+b+c \right )^{2}}\leq \frac{1+a+ca}{9}
ta chỉ cần chứng minh:
$\sum \frac{1+a+ca}{9}\leq 1 \Leftrightarrow ab+bc+ca\leq 3$
mà ta có:
$ab+bc+ca\leq \frac{\left ( a+b+c \right )^{2}}{3}$
từ đay suy ra ĐPCM "=" <=> a=b=c=1