Nesbitts Inequality

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

$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge \dfrac{3}{2}.$$

Nesbitt's inequality is a famous inequality with many unique solutions. It states that

For positive reals $a,b$ and $c,$ the following holds:

$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge \dfrac{3}{2}. \ _\square$$

Below is a list of proofs of Nesbitt's inequality. Feel free to add your own proofs.

Prove Nesbitt's Inequality

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Proof 1:

Clearing denominators and full expansion gives that it is equivalent to

$$2a^3+2b^3+2c^3\ge a^2b+a^2c+b^2a+b^2c+c^2a+c^2b.$$

However, by AM-GM

$$\dfrac{a^3+a^3+b^3}{3}\ge a^2b,$$

so summing the inequality symmetrically gives the desired inequality. $_\square$

$$$$ Proof 2:

We have

$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} = \dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ca+cb}\ge \dfrac{(a+b+c)^2}{2(ab+bc+ca)}.$$

By Titu's lemma, it remains to prove

$$\begin{array}{c}&\dfrac{(a+b+c)^2}{2(ab+bc+ca)}\ge \dfrac{3}{2} &\text{ or } &(a+b+c)^2\ge 3(ab+bc+ca), \end{array}$$

which is true after full expansion and the use of $a^2+b^2+c^2\ge ab+bc+ca$. $_\square$

$$$$ Proof 3:

Without losing generality we assume that $a\geq b\geq c$, which implies $\frac{1}{b+c}\geq\frac{1}{a+c}\geq\frac{1}{a+b}.$

Now, applying Chebyshev's inequality, we have

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

Applying Titu's lemma, we get $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\geq\frac{9}{2(a+b+c)}$, so

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

$$$$ Proof 4:

Let $S$ denote the left side of the inequality, then we can transform $S$ as follows:

$$\begin{aligned} \text{LHS} =S &=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \\ &= 1+\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}-3 \\ &=\frac{a+b+c}{b+c} +\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}-3 \\ &= \frac{1}{2}\Big[ (a+b)+(b+c)+(c+a) \Big] \left( \frac{1}{b+c}+\frac{1}{c+a} +\frac{1}{a+b} \right) -3. \end{aligned}$$

Now, let $x=a+b, y=b+c, z=c+a,$ then

$$\begin{aligned} 2S & = (x+y+z)\left(\frac{1}{y}+\frac{1}{z}+\frac{1}{x} \right) - 6 \\ & = \underbrace{\frac{x}{y}+\frac{y}{x}}_{\geq 2}+\underbrace{\frac{x}{z} + \frac{z}{x}}_{\geq2}+\underbrace{\frac{y}{z}+\frac{z}{y}}_{\geq2} - 3 \geq 3 \\ \Rightarrow S&= \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}. \ _\square \end{aligned}$$

• Titu's Lemma

• Arithmetic Mean - Geometric Mean

• Chebyshev's Inequality