Dan Sitaru's Inequality with Roots and Powers

Problem

Dan Sitaru's Inequality with Roots and Powers

Proof 1

By the AM-GM inequality, $LHS\ge 3^{10}(abc)^{4/3}.\;$ Suffice it to prove that $3^{10}(abc)^{4/3}\ge 2^43^9r^2s^2,\;$ where $r\;$ is the inradius, $s\;$ the semiperimeter of $\Delta ABC.\;$

We have a sequence of assertions equivalent to the above:

$3(abc)^4\ge 2^{12}r^6s^6,\\ 27(4Rrs)^4\ge 2^{12}r^6s^6,\\ 27R^4\ge 2^4r^2s^2,\\ \displaystyle s^2\le\frac{27R^4}{16r^2}.$

By Gerretsen's inequality, $s^2\le 4R^2+4Rr+3r^2.\;$ Thus, suffice it to prove that

$\displaystyle 4R^2+4Rr+3r^2\le\frac{27R^2}{16r^2}.$

But, with $\displaystyle t=\frac{R}{r}\ge 2,\;$ this reduces to

$\displaystyle (t-2)(27t^3+54t^2+44t+24)\ge 0,$

which is true. Equality is achieved when $R=2r,\;$ i.e., for equilateral $\Delta ABC.$

Proof 2

By the AM-GM inequality, $LHS\ge 3^{10}(abc)^{4/3}.\;$ We continue with a sequence of equivalent inequalities:

$3abc\sqrt[3]{abc}\ge 16S^2,\\ 12RS\sqrt[3]{abc}\ge 16S^2,\\ 3R\sqrt[3]{abc}\ge 4S,\\ 27R^3\cdot 4RS\ge 64S^3,\\ 3\sqrt{3}R^2\ge 4S,\\ \displaystyle S\le\frac{3\sqrt{3}R^2}{4},$

which is true. Equality is achieved when $\displaystyle S\le\frac{3\sqrt{3}R^2}{4},$ i.e., when $\Delta ABC\;$ is equilateral.

Proof 3

By the AM-GM inequality, $LHS\ge 3^{10}(abc)^{4/3}.\;$ Thus, suffice it to prove that $3(abc)^{4/3}\ge 16S^2.$

Using the expanded for of Heron's formula, the latter is equivalent to

(*)

$3(abc)^{4/3}\ge 2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4).$

Define $x=a^2,\;$ $y=b^2,\;$ $z=c^2.\;$ Then, $x,y,z\gt 0\;$ and (*) becomes

(**)

$3(xyz)^{2/3}\ge 2(xy+yz+zx)-(x^2+y^2+z^2).$

Since all term in the inequality are homogeneous functions, we may assume, WLOG, that $x+y+3=3.\;$ This implies $xy+yz+zx=3(1-t^2),\;$ where $t\in (0,1).\;$ But $2(xy+yz+zx)-(x^2+y^2+z^2)\gt 0,\;$ implying $3(1-4t^2)\ge 0,\;$ so that $\displaystyle 0\ge t\ge\frac{1}{2}\;$ and (**) becomes

(***)

$p^{2/3}\ge (1-2t)(1+2t),$

where $p=xyz.\;$ By the AM-GM inequality, $p\le 1,\;$ implying $p^{2/3}\ge p.\;$ Thus, suffice it to show that $p\ge (1-2t)(1+2t).$

BBy one of Vo Quec Ba Can's statements, $p\ge (1+t)^2(1-2t)\;$ and it is obvious that $(1+t)^2\ge 1+2t,\;$ hence (***) is true and so is (*). Equality is attained when $a=b=c,\;$ i.e., when $\Delta ABC\;$ is equilateral.

Acknowledgment

The problem has been kindly posted by Dan Sitaru at the CutTheKnotMath facebook page. Proof 1 is by Soumava Chakraborty (India), Proof 2 by Seyran Ibrahimov (Azerbaijan), Proof 3 by Leo Giugiuc (Romania).

 

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2017 Alexander Bogomolny

 62681719

Search by google: