SolFeb2506

bonus problem: show for $a, b, c > 0$ :

$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}$

Solution I For a proof by contradiction approach assume:

$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}< \frac{3}{2}$

Multiplying $a,b,c\,$ by a positive constant $t\,$ does not change the left side of the inequality. For convenience we can take $a+b+c=1\,$

$\frac{a}{b+c}+\frac{b+c}{b+c}+\frac{b}{a+c}+\frac{a+c}{a+c}+ \frac{c}{a+b}+\frac{a+b}{a+b}<3+3/2$

$\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}<\frac{9}{2}$

Next we show that this inequality cannot hold.

From the reciprocal of the harmonic-arithmetic mean inequality (http://planetmath.org/encyclopedia/ProofOfArithmeticGeometricHarmonicMeansInequality.html), we get:

$\frac{\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}}{3}\ge \frac{3}{b+c+a+c+a+b}=\frac{3}{2}$

$\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\ge\frac{9}{2}$

So the original inequality is true by proof by contradiction.

Solution II For a proof using a little calculus Define

$m=a+b+c\quad \mathrm{and}\quad f(x)=\frac{x}{m-x}\,$

Rewrite the left side of the inequality:

$\frac{a}{m-a}+\frac{b}{m-b}+\frac{c}{m-c}=f(a)+f(b)+f(c)$

Taking the second derivative of $f(x)\,$ :

${{d^2}\over{d\,x^2}}\,f\left(x\right)={{2\,x}\over{\left(m-x\right) ^3}}+{{2}\over{\left(m-x\right)^2}}>0$

Since the second derivative is positive for $0<x<m\,$ , we find that $f(x)\,$ is convex. For a convex function we can apply Jensen's (http://www.engineering.usu.edu/classes/ece/7680/lecture2/node5.html) inequality, the average of the function at points is greater than or equal to the function evaluated at the average of the points:

$\frac{f(a)+f(b)+f(c)}{3}\ge f\left(\frac{a+b+c}{3}\right)=\frac{1}{2}$

Multiplying by 3 establishes the inequality. Q.E.D.

Retrieved from "http://www.efnet-math.org/w/SolFeb2506"