1. ## absolute value

Challenge problem:

Show that: $\displaystyle \frac{|a|}{|b+c|}+\frac{|b|}{|a+c|}+\frac{|c|}{|b+ a|}\geq\frac{3}{2}$,where a,b,c are any real Nos.

$\displaystyle (|b+c|)(|a+c|)(|b+a|)\neq 0$

Moderator edit: Approved Challenge question.

2. From $\displaystyle |x+y| \le |x| + |y|$ it follows that $\displaystyle \frac{|x|}{|y+z|} \ge \frac{|x|}{|y|+|z|}$

Let $\displaystyle \alpha = |a|+|b|\:\:\:\:,\:\:\:\:\beta= |a|+|c|\:\:\:\:,\:\:\:\:\gamma= |b|+|c|$

Thus, we have:

$\displaystyle \frac{|a|}{|b+c|}+\frac{|b|}{|a+c|}+\frac{|c|}{|a+ b|} \ge \frac{1}{2}\left(\frac{\alpha+\beta-\gamma}{\gamma}+\frac{\alpha+\gamma-\beta}{\beta}+\frac{\gamma+\beta-\alpha}{\alpha}\right) =$

$\displaystyle = \frac{1}{2}\left(\frac{\alpha}{\gamma}+\frac{\gamm a}{\beta}+\frac{\beta}{\alpha}\right)+\frac{1}{2}\ left(\frac{\alpha}{\beta}+\frac{\beta}{\gamma}+\fr ac{\gamma}{\alpha}\right) - \frac{3}{2} \ge$

$\displaystyle \ge \frac{3}{2}+\frac{3}{2}-\frac{3}{2} = \frac{3}{2}$

And we're done.

On the last step we used AM-GM inequality to get

$\displaystyle \frac{x}{y}+\frac{y}{z}+\frac{z}{x} \ge 3 \sqrt[3]{\frac{x}{y} \frac{y}{z} \frac{z}{x}} = 3$

3. Originally Posted by Unbeatable0
From $\displaystyle |x+y| \le |x| + |y|$ it follows that $\displaystyle \frac{|x|}{|y+z|} \ge \frac{|x|}{|y|+|z|}$

Let $\displaystyle \alpha = |a|+|b|\:\:\:\:,\:\:\:\:\beta= |a|+|c|\:\:\:\:,\:\:\:\:\gamma= |b|+|c|$

Thus, we have:

$\displaystyle \frac{|a|}{|b+c|}+\frac{|b|}{|a+c|}+\frac{|c|}{|a+ b|} \ge \frac{1}{2}\left(\frac{\alpha+\beta-\gamma}{\gamma}+\frac{\alpha+\gamma-\beta}{\beta}+\frac{\gamma+\beta-\alpha}{\alpha}\right) =$

$\displaystyle = \frac{1}{2}\left(\frac{\alpha}{\gamma}+\frac{\gamm a}{\beta}+\frac{\beta}{\alpha}\right)+\frac{1}{2}\ left(\frac{\alpha}{\beta}+\frac{\beta}{\gamma}+\fr ac{\gamma}{\alpha}\right) - \frac{3}{2} \ge$

$\displaystyle \ge \frac{3}{2}+\frac{3}{2}-\frac{3}{2} = \frac{3}{2}$

And we're done.

On the last step we used AM-GM inequality to get

$\displaystyle \frac{x}{y}+\frac{y}{z}+\frac{z}{x} \ge 3 \sqrt[3]{\frac{x}{y} \frac{y}{z} \frac{z}{x}} = 3$
Thanks.

Another solution without using A.M-G.M would be if from:

$\displaystyle \frac{|a|}{|b+c|}+\frac{|b|}{|a+c|}+\frac{|c|}{|a+ b|} \ge \frac{1}{2}\left(\frac{\alpha+\beta-\gamma}{\gamma}+\frac{\alpha+\gamma-\beta}{\beta}+\frac{\gamma+\beta-\alpha}{\alpha}\right) =$

$\displaystyle \frac{\alpha^2+\beta^2}{2\alpha\beta}+\frac{\alpha ^2+\gamma^2}{2\alpha\gamma}+\frac{\beta^2+\gamma^2 }{2\beta\gamma}-\frac{3}{2}\geq 3-\frac{3}{2}=\frac{3}{2}$

Since :

$\displaystyle \alpha^2+\beta^2\geq 2\alpha\beta$ e.t.c e.t.c

Although my own solution is completely different.