Square Roots and Triangle Inequality

Googling for publications of Leon Bankoff I cam across of a The American Mathematical Monthly (Vol. 67, No. 1, Jan., 1960) page that, besides a problem posted by Bankoff, contained several other problems, some with solutions. One that came with two solutions had attracted my attention. The problem is pretty simple and, at first sight, admits a natural generalization. Instructively, one of the solutions is suggestive of a possible generalization, the other - while ingenious - does offer much of the insight into the essence of the problem.


Show that if $a,$ $b,$ $c$ form a triangle, then $\sqrt{a},$ $\sqrt{b},$ $\sqrt{c}$ form a triangle.

Solution 2

A necessary and sufficient condition [see the Seventh Polish Mathematical Olympiad, Prob. 9, First Round (The Mathematics Teacher, vol LI, no. 8, Dec. 1958, p. 587] that $a,$ $b,$ $c$ form a triangle is


$a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\gt (a^{4}+b^{4}+c^{4})/2.$

We must therefore show that $\sqrt{a},$ $\sqrt{b},$ $\sqrt{c}$ satisfy

$ab+bc+ca\gt (a^{2}+b^{2}+c^{2})/2.$

This readily follows since $b+c\gt a,$ $c+a\gt b,$ $a+b\gt c,$ whence $ab+ac\gt a^{2},$ $bc+ba\gt b^{2},$ $ca+cb\gt c^{2}.$

For the completeness' sake we may try solving the olympiad problem.

Let's square, say, $a\lt b+c:$ $a^{2}\lt b^{2}+c^{2}+2bc$ which is equivalent to

$\displaystyle bc\gt \frac{a^{2}-b^{2}-c^{2}}{2}.$

Choosing $a$ as the largest side, $a^{2}-b^{2}-c^{2}\gt 0,$ so that we can square the above:

$\displaystyle b^{2}c^{2}\gt \frac{a^{4}+b^{4}+c^{4}+2b^{2}c^{2}-2a^{2}c^{2}-2a^{2}b^{2}}{4}.$

Rearranging terms we get the required (1). Since all the steps in the proof are reversible, this solves the problem.

Solution 1

$\begin{align} \left|\sqrt{b}-\sqrt{c}\right|(\sqrt{b}+\sqrt{c})&\lt (\sqrt{a})^{2}\\ &\lt b+c\\ &\lt (\sqrt{b}+\sqrt{c})^{2}. \end{align}$

It follows that $\left|\sqrt{b}-\sqrt{c}\right|\lt\sqrt{a}\lt\sqrt{b}+\sqrt{c}.$

This solution suggests a generalization:


Let an integer $n\gt 1.$ Show that if $a,$ $b,$ $c$ form a triangle, then $\sqrt[n]{a},$ $\sqrt[n]{b},$ $\sqrt[n]{c}$ form a triangle.

Here's the solution by Wojtek Wawrów posted at the CutTheKnotMath facebook page.

$\begin{align} (\sqrt[n]{a})^{n}&= a\\ &\lt b+c\\ &=(\sqrt[n]{b})^{n}+(\sqrt[n]{c})^{n}\\ &\lt (\sqrt[n]{b})^{n}+(\sqrt[n]{c})^{n}+\displaystyle\sum_{k=1}^{n-1}{n\choose k}(\sqrt[n]{b})^{k}(\sqrt[n]{c})^{n-k}\\ &=(\sqrt[n]{b}+\sqrt[n]{c})^{n}. \end{align}$

Therefore, $\sqrt[n]{a}\lt\sqrt[n]{b}+\sqrt[n]{c}.$

Note: This proof suggests a further generalization: Let $f:\,\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ be 1-1 and such that $f(a+b)\lt f(a)+f(b).$ Then, if $a,b,c$ form a triangle, then so do $f^{-1}(a),f^{-1}(b),f^{-1}(c).$

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