Áp dụng bđt $Bunhiacopxki:S^2=(a_1+a_2+...+a_{2015})^2 \le 2015(a_1^2+a_2^2+...a_{2015}^2)$$\Rightarrow a_1^2+a_2^2+...a_{2015}^2 \ge \frac{S^2}{2015}$Áp dụng bđt $Cauchy-Schwarz$, ta có :$VT=\frac{a_1^2}{a_1.S-a_1^2}+\frac{a_2^2}{a_2.S-a_2^2}+...+\frac{a_{2015}^2}{a_{2015}.S-a_{2015}^2} \ge\frac{(a_1+a_2+...+a_n)^2}{(a_1+a_2+...+a_n).S-(a_1^2+a_2^2+...+a_{2015}^2)}$$ \ge \frac{S^2}{S^2-\frac{S^2}{2015}}= \frac{2015}{2014}$
Áp dụng bđt $Bunhiacopxki:S^2=(a_1+a_2+...+a_{2015})^2 \ge 2015(a_1^2+a_2^2+...a_{2015}^2)$$\Rightarrow a_1^2+a_2^2+...a_{2015}^2 \le \frac{S^2}{2015}$Áp dụng bđt $Cauchy-Schwarz$, ta có :$VT=\frac{a_1^2}{a_1.S-a_1^2}+\frac{a_2^2}{a_2.S-a_2^2}+...+\frac{a_{2015}^2}{a_{2015}.S-a_{2015}^2} \ge\frac{(a_1+a_2+...+a_n)^2}{(a_1+a_2+...+a_n).S-(a_1^2+a_2^2+...+a_{2015}^2)}$$ \ge \frac{S^2}{S^2-\frac{S^2}{2015}}= \frac{2015}{2014}$
Áp dụng bđt $Bunhiacopxki:S^2=(a_1+a_2+...+a_{2015})^2 \
le 2015(a_1^2+a_2^2+...a_{2015}^2)$$\Rightarrow a_1^2+a_2^2+...a_{2015}^2 \
ge \frac{S^2}{2015}$Áp dụng bđt $Cauchy-Schwarz$, ta có :$VT=\frac{a_1^2}{a_1.S-a_1^2}+\frac{a_2^2}{a_2.S-a_2^2}+...+\frac{a_{2015}^2}{a_{2015}.S-a_{2015}^2} \ge\frac{(a_1+a_2+...+a_n)^2}{(a_1+a_2+...+a_n).S-(a_1^2+a_2^2+...+a_{2015}^2)}$$ \ge \frac{S^2}{S^2-\frac{S^2}{2015}}= \frac{2015}{2014}$