absolute value

**alexandros** · June 9th 2010, 05:54 AM

Challenge problem:

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

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

Moderator edit: Approved Challenge question.

**Unbeatable0** · June 10th 2010, 01:55 AM

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

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

Thus, we have:

$ \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) = $

$ = \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 $

$ \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

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

**alexandros** · June 10th 2010, 05:35 AM

Originally Posted by Unbeatable0

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

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

Thus, we have:

$ \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) = $

$ = \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 $

$ \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

$ \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:

$ \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) = $

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

Since :

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

Although my own solution is completely different.

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