Nesbitt's inequality

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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:



First proof[edit]

Starting from Nesbitt's inequality(1903)


we transform the left hand side:


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[edit]

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

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


\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

Addition yields Nesbitt's inequality.

Third proof[edit]

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[edit]

Starting from Nesbitt's inequality (1903)


We add 3 to both sides.


Now this can be transformed into:


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[edit]

Starting from Nesbitt's inequality (1903)


we substitute :a+b=x, b+c=y, c+a=z.

Now, we get


this can be transformed to


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

Sixth proof[edit]

We have


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


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.


External links[edit]