$ab + bc + ca \le aa + bb + cc$

Problem

The inequality below, though simple, is useful and often comes up in proofs of more involved inequalities. Here, I'd like to document several of its proofs.

Proof 0

This proof is illustrated by the applet above. It was suggested by Teo López Puccio.

The bottom segment is split into three of lengths $a,\,$ $b,\,$ and $c.\,$ We assume $a\ge b\ge c.\,$ For the value of the slider $u=1,\,$ the total colored area equals $ab+bc+ca.\,$ When the value of the slider $u=0,\;$ the total colored area equals $a^2+b^2+c^2.\,$ One can observe that the former could not exceed the latter.

Proof 1

By the AM-GM inequality, $\displaystyle xy\le\frac{x^2+y^2}{2}.\;$ So we have

\displaystyle\begin{align} ab+bc+ca &\le \left(\frac{a^2+b^2}{2}\right)+\left(\frac{b^2+c^2}{2}\right)+\left(\frac{c^2+a^2}{2}\right)\\ &=a^2+b^2+c^2. \end{align}

Proof 2

WLOG, we may assume, say, $a\le b\le c.\;$ Then by the Rearrangement Inequality,

$ab + bc + ca \le aa + bb + cc=a^2 + b^2 + c^2.$

Proof 3

Observe the obvious $(a-b)^2+(b-c)^2+(c-a)^2\ge 0.\;$ This is equivalent to

$2(a^2+b^2+c^2)\ge 2(ab + bc + ca).$

Proof 4

$\displaystyle\sum_{cycl}a^2-\sum_{cycl}ab=\frac{1}{2}\sum_{cycl}(a-b)^2\ge 0.$

Proof 5

The function $\displaystyle f(a,b,c)=\sum_{cycl}a^2-\sum_{cycl}ab\;$ is homogeneous of degree $2:\;$ $f(ta,tb,tc)=t^2f(a,b,c),\;$ for $t\ne 0.\;$ Thus, we may apply various normalizations, e.g., $a=1,\;$ $b=1+x,\;$ $c=1+y.\;$ The substitution transforms the inequality into

$(1+x)+(1+x)(1+y)+(1+y)\le 1+(1+x)^2+(1+y)^2,$

which reduces to

$(1+x)+(1+x)(1+y)+(1+y)\le 1+(1+x)^2+(1+y)^2,$

and, subsequently, to $xy\le x^2+y^2.$ But from $(x-y)^2\ge 0\;$ it follows that already $\displaystyle xy\le\frac{1}{2}(x^2+y^2)\;$ which is stronger than $xy\le x^2+y^2.$

Proof 6

Reaching $xy\le x^2+y^2$ as in the previous proof, we observe that this is equivalent to $\displaystyle 0\le\left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4}.$

Proof 7

Let's denote $p=a+b+c\;$ and $q=ab+bc+ca.\;$ $a,b,c\;$ are the roots of a third-degree polynomial $f(t)=t^3-pt^2+qt-abc.\;$ Since this polynomial has three real roots $a,b,c,\;$ its derivative $f'(t)=3t^2-2pt+q\;$ has two real roots, meaning that its discriminant is not negative: $p^2-3q\ge 0.$ In terms of $a,b,c\;$ this is exactly the required inequality.

Proof 8

Let's consider a second degree polynomial $f(x)=x^2-x(b+c)+(b^2+c^2-bc).\;$ The discriminant of this polynomial,

$D=(b+c)^2-4(b^2+c^2-bc)=-3(b-c)^2\lt 0,$

meaning that $f(t)\;$ has no real roots (unless $b=c)\;$ and, consequently, is of the same sign for all $t\in\mathbb{R}.\;$ Since it is obviously positive for large values of $t,\;$ it is positive everywhere. In particular, if $b\ne c,\;$ $f(a)\gt 0.\;$ From the quadratic formula $f(a)=0,\;$ iff $a=b=c.$

Proof 9

This proof has been communicated by Leo Giugiuc.

By the Ionescu-Weitzenbock inequality, $x^2+y^2+z^2\ge 4\sqrt{3}S,\;$ where $S=[\Delta ABC],\;$ the area of an acute triangle $ABC\;$ and $x,y,z\;$ its side lengths.

Let $x^2=b^2+c^2,\;$ $y^2=c^2+a^2,\;$ $z^2=a^2+b^2,\;$ where $a,b,c\;$ are real numbers. $2S=\sqrt{a^2b^2+b^2c^2+c^2a^2}.\;$ So we have

$a^2+b^2+c^2\ge \sqrt{3(a^2b^2+b^2c^2+c^2a^2)}.\;$

But

\begin{align} \sqrt{3(a^2b^2+b^2c^2+c^2a^2)}&\ge|ab|+|bc|+|ca|\\ &\ge ab+bc+ca.\;\end{align}

Proof 10

This illustration is by Nassim Nicholas Taleb.

Graphical representation: ridge a=b for different values of c. Can't get a clean RegionPlot3D.

Finally, as we converge to equality from below.

Proof 11

Kunihiko Chikaya posted the following proof with the comment: "Here is the solution to Japanese Kids at age of 13, 14, 15."

\begin{align} f(a) &= a^2 + b^2 + c^2 - ab - bc - ca\\ &= a^2 - (b + c ) a + b^2 - bc + c^2\\ &= [a - (b + c) / 2] ^2 + (3/4)\cdot (b-c)^2\\ &\ge 0. \end{align}

Equality holds if only if $a - (b + c) / 2 = 0\;$ and $b - c = 0,\;$ $a = b = c.$

Proof 12

Sam Walters has observed that the inequality is a one step application of the Cauchy-Schwarz inequality:

$ab+bc+ca \le \sqrt{a^2+b^2+c^2}\cdot\sqrt{b^2+c^2+a^2}=a^2+b^2+c^2.$

Illustration

Here's an illustration by Gary Davis.

References

1. A. Engel, Problem-Solving Strategies, Springer Verlag, 1998