Xét: $\frac{ab}{a^{2}+b^{2}}=\frac{\frac{1}{2}(a^{2}+b^{2})}{a^{2}+b^{2}}-\frac{\frac{1}{2}(a-b)^{2}}{a^{2}+b^{2}}=\frac{1}{2}-\frac{\frac{1}{2}(a-b)^{2}}{a^{2}+b^{2}}\leq \frac{1}{2}$ (1)
Tương tự:$\frac{bc}{b^{2}+c^{2}}\leq \frac{1}{2};\frac{ac}{c^{2}+a^{2}}\leq \frac{1}{2}$(2)
Xét: $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=3+(\frac{a}{b}+\frac{b}{a})+(\frac{b}{c}+\frac{c}{b})+(\frac{a}{c}+\frac{c}{a})$
Áp dụng BĐT Cauchy ta có:
$\frac{a}{b}+\frac{b}{a}\geq2$ tượng tự $\frac{b}{c}+\frac{c}{b}\geq 2;\frac{a}{c}+\frac{c}{a}\geq 2$
$\Rightarrow (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9\Leftrightarrow \frac{1}{4}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq \frac{9}{4}(3)$
Từ (1);(2);(3) $\Rightarrow A\geq \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{9}{4}=\frac{15}{4}$
Xong òi....Chiuu...cho mình xin lỗi nhá đối với những gì mình đã hiểu nhầm....Sorry.....
JUST WANNA SAY SORRY