An Area Inequality

An inequality that is proved by resorting to a geometric interpretation

Solution

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There may be many ways to prove the required inequality, but one starts with a recognition of the expression $av - bu\;$ as the area of the parallelogram formed by vectors $(a, b)\;$ and $(u, v):\;$

parallelogram formed by two vectors

How does this help? Let $A = a^{2} + b^{2} + u^{2} + v^{2} + au + bv.\;$ Then

$2A = (a + u)^{2} + (b + v)^{2} + a^{2} + b^{2} + u^{2} + v^{2}.$

Consider one of the triangles in the diagram. Denote its sides $X, Y, Z\;$ so that, by the Pythagorean theorem

$\begin{align} X^{2} &= a^{2} + b^{2},\\ Y^{2} &= u^{2} + v^{2},\\ Z^{2} &= (a + u)^{2} + (b + v)^{2}, \end{align}$

and $2A = X^{2} + Y^{2} + Z^{2}.$

Now, we come to a

Lemma

In a triangle with sides $X, Y, Z,\;$ area $S\;$ satisfies

$\displaystyle S \le \frac{X^{2} + Y^{2} + Z^{2}}{6}.$

Proof

Let $α\;$ be the angle formed by $X\;$ and $Y.\;$ Then

$\displaystyle S = \frac{X·Y·\sinα}{2} = \frac{2X·Y·\sinα}{4} \le \frac{(X - Y)^{2} + 2XY}{4} = \frac{X^{2} + Y^{2}}{4}.$

(This is so simply because, if $α\;$ is an angle in a triangle then, $0 < \sinα \le 1,\;$ and $0 \le (X - Y)^{2}.)\;$

Similar inequalities hold for pairs $X, Z\;$ and $Y, Z:$

$\displaystyle\begin{align} S &\le (X^{2} + Y^{2}) / 4,\\ S &\le (Y^{2} + Z^{2}) / 4,\\ S &\le (Z^{2} + X^{2}) / 4. \end{align}$

Adding the three proves the lemma.

Proof of the inequality

Let T be the area of the parallelogram in the diagram. From Lemma,

$\displaystyle\begin{align} T/2 &\le \frac{X^{2} + Y^{2} + Z^{2}}{6}, \text{i.e.,}\\ T &\le \frac{X^{2} + Y^{2} + Z^{2}}{3}. \end{align}$

But, as the problem stipulates, $T = 1,\;$ so that

$X^{2} + Y^{2} + Z^{2} \ge 3.$

But

$\displaystyle\begin{align} X^{2} + Y^{2} + Z^{2} &= 2A\\ &= 2(a^{2} + b^{2} + u^{2} + v^{2} + au + bv), \end{align}$

implying

$\displaystyle a^{2} + b^{2} + u^{2} + v^{2} + au + bv \ge \frac{3}{2}.$

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Copyright © 1996-2017 Alexander Bogomolny

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