Đáp án:
$\frac{a}{b+c}+ \frac{b}{c+a}+\frac{c}{a + b} < 2$
Giải thích các bước giải:
$Với$ $a,b,c$ $là$ $3$ $cạnh$ $của$ $1$ $tam$ $giác$ $ta$ $có$:
$\frac{a}{b + c} < \frac{a + a}{a + b + c} = \frac{2a}{a+b+c} (1)$
$\frac{b}{c + a} < \frac{b + b}{a + b + c} =\frac{2b}{a + b + c} (2)$
$\frac{c}{a + b} < \frac{c + c}{a + b + c} =\frac{2c}{a + b + c} (3)$
$Cộng$ $vế$ $với$ $vế$ $của$ $(1) , (2) , (3)$ $ta$ $có :$
$\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} < \frac{2a}{a + b + c} + \frac{2b}{a + b + c} +\frac{2c}{a + b + c}$
$→ \frac{a}{b+c}+ \frac{b}{c+a}+\frac{c}{a + b} < \frac{2a + 2b + 2c}{a + b + c}$
$→ \frac{a}{b+c}+ \frac{b}{c+a}+\frac{c}{a + b} < \frac{2(a + b + c)}{a + b + c}$
$→ \frac{a}{b+c}+ \frac{b}{c+a}+\frac{c}{a + b} <2$
