c) BDT này không khó $bdt\Leftrightarrow \frac{1}{\left(1+\dfrac{b}{a} \right)^2}+\frac{1}{\left(1+\dfrac{c}{a} \right)^2} \ge \frac{1}{1+\dfrac{b.c}{a.a}}$Đặt $\frac ba=x,\frac ca=y$$VT=\frac{1}{(x+1)^2}+\frac{1}{(y+1)^2} \ge \frac{1}{2(x^2+1)}+\frac{1}{2(y^2+1)} \ge \frac{1}{xy+1}$
c) BDT này không khó $bdt\Leftrightarrow \frac{1}{\left(1+\dfrac{b}{a}^2 \right)}+\frac{1}{\left(1+\dfrac{c}{a}^2 \right)} \ge \frac{1}{1+\dfrac{b.c}{a.a}}$Đặt $\frac ba=x,\frac ca=y$$VT=\frac{1}{(x+1)^2}+\frac{1}{(y+1)^2} \ge \frac{1}{2(x^2+1)}+\frac{1}{2(y^2+1)} \ge \frac{1}{xy+1}$
c) BDT này không khó $bdt\Leftrightarrow \frac{1}{\left(1+\dfrac{b}{a} \right)
^2}+\frac{1}{\left(1+\dfrac{c}{a} \right)
^2} \ge \frac{1}{1+\dfrac{b.c}{a.a}}$Đặt $\frac ba=x,\frac ca=y$$VT=\frac{1}{(x+1)^2}+\frac{1}{(y+1)^2} \ge \frac{1}{2(x^2+1)}+\frac{1}{2(y^2+1)} \ge \frac{1}{xy+1}$