Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$

Question

I am currently attempting to prove the following inequality

$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$ for all $ a,b,c>0$

My instinctive plan of attack is to use the AM/GM inequality with $x_1=\dfrac{a}{b+c}$ etc.

Using that I get this

$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}}$

From here, I used the fact that $(a+b)(b+c)(a+c)\geq 8abc$, which can be easily proven by considering that $a+b\geq 2\sqrt{ab}$

But by using this, I get the following...

$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}} \leq 3 \times \sqrt[3]{\dfrac{abc}{8abc}} = \dfrac{3}{2}$

Everything seems so perfect because I get the value $\dfrac{3}{2}$ as required, but this method isn't valid due to the change in direction! What is going on?

Is there a way of proving this inequality otherwise then?

Answer 1

Use AM > HM on $\frac{a+b+c}{a+b}, \frac{a+b+c}{b+c}$ and $\frac{a+b+c}{c+a}$.

Answer 2

If we denote $s:=a+b+c$, we have to minimize $$\frac{a}{s-a}+\frac{b}{s-b}+\frac{c}{s-c}$$ for $a,b,c>0$.

Using new variables $x:=a/s$, $y:=b/s$, $z:=c/s$ we can see, that this is equivalent to minimizing $$\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z}$$ for $x+y+z=1$.

If we notice that the function $$f(x)=\frac{x}{1-x}=-1+\frac1{1-x}=-1-\frac1{x-1}$$ is convex on the interval $(0,1)$, we have $$\frac{f(x)+f(y)+f(z)}3 \ge f\left(\frac{x+y+z}3\right),$$ i.e. $$f(x)+f(y)+f(z) \ge 3f(1/3).$$ By computing $f(1/3)=1/2$ we see, that the last inequality is precisely $$\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z} \ge \frac32.$$

Answer 3

By Cauchy-Schwarz $$\begin{align}\sum_{cyc}\frac{a}{b+c}&=\sum_{cyc}\left(\frac{a}{b+c}+1\right)-3\\ &=\sum_{cyc}a\sum_{cyc}\frac{1}{b+c}-3\\ &=\frac{1}{2}\sum_{cyc}(b+c)\sum_{cyc}\frac{1}{b+c}-3\\ &\geq\frac{1}{2}\cdot9-3=\frac{3}{2}.\end{align}$$ Done!

Another way.

By C-S $$\sum\limits_{cyc}\frac{a}{b+c}=\sum\limits_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(ab+ac)}=\frac{(a+b+c)^2}{2(ab+ac+bc)}\geq\frac{3}{2},$$ where the last inequality it's just $\sum\limits_{cyc}(a-b)^2\geq0$. Done!

Another way. $$\sum\limits_{cyc}\frac{a}{b+c}-\frac{3}{2}=\sum\limits_{cyc}\left(\frac{a}{b+c}-\frac{1}{2}\right)=\sum\limits_{cyc}\frac{2a-b-c}{2(b+c)}=\sum\limits_{cyc}\frac{a-b-(c-a)}{2(b+c)}=$$ $$=\sum\limits_{cyc}(a-b)\left(\frac{1}{2(b+c)}-\frac{1}{2(a+c)}\right)=\sum\limits_{cyc}\frac{(a-b)^2}{2(a+c)(b+c)}\geq0.$$ Done!

Answer 4

Here is another nice proof.

We have to prove;

$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{3}{2}\\\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\geq \frac{3}{2}+3\\(a+b+c)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\geq \frac{9}{2}\\([b+c]+[a+c]+[a+b])\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\geq 9\\$$Now we use this well known inequality for $x>0\to x+\frac{1}{x}\geq2$;

(quick proof: $\left (\sqrt x-\frac{1}{\sqrt x}\right )^2\geq0\to x+\frac1x\geq2$) $$1+1+1+\left[\frac{a+b}{a+c}+\frac{a+c}{a+b}\right]+\left[\frac{a+b}{b+c}+\frac{b+c}{a+b}\right]+\left[\frac{a+c}{b+c}+\frac{b+c}{a+c}\right]\geq 9\\q.e.d.$$

Answer 5

We can also use rearrangement inequality to prove it.

$\displaystyle\frac a{b+c}+\frac b{a+c}+\frac c{a+b}$
$\displaystyle=\frac12\left(\left(\frac a{b+c}+\frac b{a+c}+\frac c{a+b}\right)+\left(\frac a{b+c}+\frac b{a+c}+\frac c{a+b}\right)\right)$
$\displaystyle\geq\frac12\left(\left(\frac a{a+c}+\frac b{a+b}+\frac c{b+c}\right)+\left(\frac a{a+b}+\frac b{b+c}+\frac c{a+c}\right)\right)$
$=\dfrac32$

Answer 6

\begin{eqnarray*} % \nonumber to remove numbering (before each equation) && \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}-\frac{3}{2}\\ &=& \frac{2(x^3+y^3+z^3)-(x^2y+xy^2+y^2z+yz^2+z^2x+zx^2)}{2 (x + y) (x + z) (y + z)}\\ &=& \frac{(x+y)(x-y)^2+(y+z)(y-z)^2+(z+x)(z-x)^2}{2 (x + y) (x + z) (y + z)}\geq0 \end{eqnarray*} with equality holds if and only if $x=y=z$.