# Nesbitt's inequality

In mathematics, Nesbitt's inequality is a special case of the Shapiro inequality. It states that for positive real numbers a, b and c we have:

$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}.$

## Proof

### First proof

Starting from Nesbitt's inequality(1903)

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

we transform the left hand side:

$\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3\geq\frac{3}{2}.$

Now this can be transformed into:

$((a+b)+(a+c)+(b+c))\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\geq 9.$

Division by 3 and the right factor yields:

$\frac{(a+b)+(a+c)+(b+c)}{3}\geq\frac{3}{\frac{1}{a+b}+\frac{1}{a+c}+ \frac{1}{b+c}}.$

Now on the left we have the arithmetic mean and on the right the harmonic mean, so this inequality is true.

We might also want to try to use GM for three variables.

### Second proof

Suppose $a \ge b \ge c$, we have that

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

define

$\vec x = (a, b, c)$
$\vec y = \left(\frac 1 {b+c} , \frac 1 {a+c} , \frac 1 {a+b}\right)$

The scalar product of the two sequences is maximum because of the Rearrangement inequality if they are arranged the same way, call $\vec y_1$ and $\vec y_2$ the vector $\vec y$ shifted by one and by two, we have:

$\vec x \cdot \vec y \ge \vec x \cdot \vec y_1$
$\vec x \cdot \vec y \ge \vec x \cdot \vec y_2$

### Third proof

The following identity is true for all $a,b,c:$

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

This clearly proves that the left side is no less than $\frac{3}{2}$ for positive a,b and c.

Note: every rational inequality can be solved by transforming it to the appropriate identity, see Hilbert's seventeenth problem.

### Fourth proof

Starting from Nesbitt's inequality (1903)

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

We add $3$ to both sides.

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

Now this can be transformed into:

$(a+b+c)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\geq\frac{9}{2}$

Multiply by $2$ on both sides.

$((b+c)+(a+c)+(a+b))\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\geq 9$

Which is true by the Cauchy-Schwarz inequality.

### Fifth proof

Starting from Nesbitt's inequality (1903)

$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}$,

we substitute :$a+b=x, b+c=y, c+a=z$.

Now, we get

$\frac{x+z-y}{2y}+\frac{y+z-x}{2x}+\frac{x+y-z}{2z}\geq\frac{3}{2};$

this can be transformed to

$\frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}\geq\frac{6}{1}$

which is true, by inequality of arithmetic and geometric means.

### Sixth proof

We have

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

Multiplying the first fraction by $\frac{a}{a}$, the second by $\frac{b}{b}$, and the third by $\frac{c}{c}$ gives

$\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\geq\frac{3}{2}$

Applying Titu's Lemma we get

$\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\geq \frac{(a+b+c)^2}{2(ab+bc+ca)}\geq \frac{3}{2}$

Thus it suffices to prove that

$(a+b+c)^2\geq 3(ab+bc+ca)$

But expanding and rearranging we arrive at

$a^2+b^2+c^2-ab-bc-ca\geq 0$

Which can be written as

$\frac{1}{2}((a-b)^2+(b-c)^2+(c-a)^2)\geq 0$

which is true.