An Inequality with a Generic Proof


Dorin Marghidanu has kindly posted a problem with solution at the CutTheKnotMath facebook page. The method of solution points naturally to a generalization.

$x,y,z,a,b\ge 0,\;$ with $x+y+z=S,\;$ prove the inequality

$2\sqrt{b}+\sqrt{aS+b}\le \sqrt{ax+b}+ \sqrt{ay+b}+ \sqrt{az+b}\le \sqrt{3aS+9b}.$


$x,y,p\ge 0,\;$ with $x+y+z=S,\;$ the following inequality holds

$\sqrt{p}+\sqrt{p+x+y}\le \sqrt{p+x}+ \sqrt{p+y}.$

Indeed, by squaring, one obtains

$p+(p+x+y)+2\sqrt{p(p+x+y)}\le (p+x)+(p+y)+ 2\sqrt{(p+x)(p+y)}$

which is equivalent to $xy\ge 0.$ The equality only occurs when either $x=0\;$ or $y=0,\;$ i.e., for the pairs $(0,k)\;$ and $(k,0),\;$ with $k\ge 0.$


We start with the left inequality:

$\begin{align} (\sqrt{ax+b}+ \sqrt{ay+b})+ \sqrt{az+b} &\ge (\sqrt{b}+ \sqrt{ax+ay+b})+ \sqrt{az+b}\\ &=\sqrt{b} +(\sqrt{ax+ay+b})+ \sqrt{az+b})\\ &\ge \sqrt{b}+ (\sqrt{b}+\sqrt{ax+ay+az+b})\\ &=2\sqrt{b}+\sqrt{aS+b}. \end{align}$

The right inequality follows in one step from either QM-AM inequality $\displaystyle\frac{u+v+w}{3}\le\sqrt{\frac{u^2+v^2+w^2}{3}},$ or, given that $t=\sqrt{s}\;$ is a concave function, from Jensen's inequality $\displaystyle\frac{f(u)+f(v)+f(w)}{3}\le f\left(\frac{u+v+w}{3}\right).$ Both ways lead to

$\displaystyle \sqrt{ax+b}+ \sqrt{ay+b}+ \sqrt{az+b}\le 3\sqrt{\frac{(ax+b)+(ay+b)+(az+b)}{3}}=\sqrt{3aS+9b}.$

The equality only holds when $\sqrt{ax+b}+ \sqrt{ay+b}+ \sqrt{az+b},\;$ i.e., when $x=y=z.$


The problem and the solution by Professor Marghidanu are of the remarkable kind where a formulation and a proof of a particular case reveal all the trappings of a general formulation and its proof. Problems like that help acclimate the student to working with general case by taking a few steps from one particular case to the next. Examples like that are known as generic proofs, the proofs of special cases that nonetheless included all the elements of the general statement. Generic proofs have been extensively written about, see, e.g., [U. Leron and O. Zaslavsky].

The general formulation is suggestive:

An Inequality with a Generic Proof


  1. U. Leron and O. Zaslavsky, Generic Proofs: Reflections on Scope and Method, in The Best Writing on Mathematics 2014 (Mircea Pitici, ed.), Princeton University Press, 2014


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