# Inequality. $\frac{ab+c}{a+b}+\frac{ac+b}{a+c}+\frac{bc+a}{b+c} \geq 2.$

Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that (using rearrangements inequalities, you can also view this exercise here, exercise number 3.1.8 )

$$\frac{ab+c}{a+b}+\frac{ac+b}{a+c}+\frac{bc+a}{b+c} \geq 2.$$

thanks.

-
@AlexBecker. I try using $\displaystyle \left(\frac{1}{b+c}, \frac{1}{a+c}, \frac{1}{a+b}\right)$ & $(a,b,c)$ and I suppose that $a \leq b \leq c$ and then when I applied the rearrangements inequality I added $\displaystyle \frac{bc}{b+c}+\frac{ab}{a+b}+\frac{ac}{a+c}$. –  Iuli Sep 5 '12 at 20:13

## 4 Answers

Observe $a + b = 1 - c$, $a + c = 1 - b$, and $b + c = 1 - a$, so the desired inequality is $$\frac{ab+c}{1 - c}+\frac{ac+b}{1 - b}+\frac{bc+a}{1 - a} \geq 2$$ Similarly, we substitute $c = 1 - a - b$, $b = 1 - a - c$, and $a = 1 - b - c$ in the numerator, and the desired inequality becomes $$\frac{ab + 1 - a - b }{1 - c}+\frac{ac+1 - a - c}{1 - b}+\frac{bc+ 1 - b - c}{1 - a} \geq 2$$ This can be rewritten as $$\frac{(1 - a)(1-b)}{1 - c}+\frac{(1 - a)(1 - c)}{1 - b}+\frac{(1 - b)(1 - c)}{1 - a} \geq 2$$ It's natural to let $A = 1 - a$, $B = 1 - b$, and $C = 1 - c$ here. So we want to show under the condition that $A + B + C = 2$ that we have the following. $$\frac{AB}{C}+\frac{AC}{B}+\frac{BC}{A} \geq 2 {\hspace 1 in}(*)$$ Without loss of generality, we may assume $A \leq B \leq C$. Then the rearrangement inequality says the left-hand side of $(*)$ is at least as large of what you get by any permutation of the denominator. So you have $$\frac{AB}{C}+\frac{AC}{B}+\frac{BC}{A} \geq \frac{AB}{A}+\frac{AC}{C}+\frac{BC}{B}$$ $$= B + A + C$$ $$= 2$$

-

Using the identity $\displaystyle\frac{ab+c}{a+b}=\frac{ab+c^2}{a+b}+c,$ it suffices to check that $$\sum_{cyc}\frac{ab}{a+b}+\sum_{cyc}\frac{c^2}{a+b}\geq a+b+c.$$ Note that the sequences $\{a^2,b^2,c^2\}$ and $\left\{\dfrac{1}{b+c},\dfrac{1}{c+a},\dfrac{1}{a+b}\right\}$ are similarly sorted, so that we obtain $$\sum_{cyc}\frac{c^2}{a+b}\geq\sum_{cyc}\frac{a^2}{a+b},$$ Which, in accordance with $\dfrac{ab}{a+b}+\dfrac{a^2}{a+b}=a,$ leads us to our desired result. Equality occurs in the original inequality if and only if $a=b=c.$ $\Box$

-

One approach (which is probably not what you mean by «using rearrangements inequalities»...) is to find the minimum of the left hand side of your inequality subject to the condition that $a+b+c=1$, using the method of Lagrange multipliers. A straightforward computation —most importantly, a very uneventlful one!— shows there is a unique extreme point, which has to be a minimum, and evaluating there shows that the extreme value is $2$.

We invented computers to do this sort of thing for us: using Mathematica, I get

In[27]:= f = (a b + c)/(a + b) + (a c + b)/(a + c) + (b c + a)/(b + c);

In[28]:= sol = Solve[
{D[f, a] == k, D[f, b] == k, D[f, c] == k, a + b + c == 1, a > 0,
b > 0, c > 0},
{a, b, c, k}
]

Out[28]= {{a -> 1/3, b -> 1/3, c -> 1/3, k -> 1/2}}

In[29]:= f /. sol[[1]]

Out[29]= 2

-
I wonder what percentage of the inequalities in the link in the question can be obtained by the same approach, simply doing minimization/maximization using Lagrange multipliers... –  Mariano Suárez-Alvarez Sep 5 '12 at 20:18

The inequality follows $$\frac{(b+c)(c+a)}{a+b}+\frac{(a+c)(a+b)}{b+c}+\frac{(a+b)(b+c)}{a+c}\geq 2$$ Let $a+b=z,b+c=x,c+a=y$ then $x+y+z=2$. Therefore $$x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}\geq xyz(x+y+z)$$ This is true by AM-GM.

P/s: Sorry for my bad English.

-