An Asymmetric Inequality

The following problem was offered at the 1992 Brazilian National Olympiad:

an asymmetric ineqaulity


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Copyright © 1996-2018 Alexander Bogomolny

Prove, for all positive numbers $a,b,c,\;$ the following inequality holds:

$(a+b)(a+c)\ge 2\sqrt{abc(a+b+c)}.$

Proof 1

Quantities $a+b,\;$ $b+c,\;$ and $c+a\;$ satisfy the triangle inequalities and, hence can be thought of as the side length of $\Delta ABC:\;$ $AB=a+b,\;$ $BC=b+c,\;$ and $AC=c+a.\;$

In terms of $a,b,c,\;$ the semiperimeter $p=a+b+c\;$ and, according to Heron's formula, $[\Delta ABC]=\sqrt{abc(a+b+c)},\;$ where $[\Delta ABC]\;$ is the area of $\Delta ABC.$

On the other hand, $[\Delta ABC]=\frac{1}{2}AB\cdot AC\cdot\sin\angle BAC\ge\frac{1}{2}AB\cdot AC,\;$ which immediately implies $\frac{1}{2}AB\cdot AC\le\sqrt{abc(a+b+c)},\;$ or, $(a+b)(a+c)\ge 2\sqrt{abc(a+b+c)}.$

Proof 2

Set $x=a(a+b+c),\;$ $y=bc.\;$ Then $x+y = (a+b)(a+c)\;$ and $\displaystyle\frac{x+y}{2}\ge\sqrt{xy}=\sqrt{abc(a+b+c)}.$

Proof 3

Observe that $(a+b)^2(a+c)^2-4abc(a+b+c)=(a^2+ab+ac-bc)^2.$


The problem with the proof (Proof 1) by Leo Giugiuc and Dan Sitaru has been posted by Leo Giugiuc at the CutTheKnotMath facebook page. Proof 2 is by Alexander Price; Proof 3 is by Imad Zak.

Dorin Marghidanu has made the following observation:

If $a,b,c$ satisfy the equality in $(a+b)(a+c)\ge 2\sqrt{abc(a+b+c)}\;$ then necessarily $a\lt\min\{b,c\}.$

Indeed, the equality holds iff $a(a+b+c)=bc,\;$ i.e., iff $a(a+b)=c(b-a)\;$ and $a(a+c)=b(c-a).\;$ Since the left-hand sides are positive, so are the right-hand sides, implying $a\lt b\;$ and $a\lt c.$

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Copyright © 1996-2018 Alexander Bogomolny


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