Inequality (2)

Algebra Level pending

For positive reals $a,b,c,d$ , the following is true for a positive $n$ ,

$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge n$

then $n=?$

The answer is 2.

1 solution

Mohammed Imran
Apr 4, 2020

Let $a \leq b \leq c \leq d$ . By Rearrangement Inequality, we have $\sum_{cyc} \frac{a}{b+c} \geq \sum_{cyc} \frac{a}{a+b}$ and $\sum_{cyc} \frac{a}{b+c} \geq \sum_{cyc} \frac{b}{a+b}$ when adding both the inequalities, we have $2\sum_{cyc} \frac{a}{b+c} \geq \sum_{cyc} \frac{a+b}{a+b}=4$ and hence we have $n=\frac{4}{2}=\boxed{2}$

Inequality (2)

The answer is 2.

1 solution

0 pending reports