RomanoNorwegian Inequality


Here is a sample inequality from a recent book 300 Romanian Mathematical Challenges by Professor Radu Gologan, Daniel Sitaru and Leonard Giugiuc. The problem is an invention of Lorian Nelu Saceanu, Norway - Romania. Solution 1 below is by Daniel Sitaru and Leonard Giugiuc, Solution 2 is by Grégoire Nicollier.

RomanoNorwegian Inequality

Solution 1

Denote $x=\cot A,\;$ $y=\cot B,\;$ $z=\cot C.\;$ Then $x,y,z\ge 0\;$ and $xy+yz+zx=1.\;$ We need to prove $\sqrt {x}+\sqrt{y}+\sqrt{z}\ge 2.$

WLOG, let's assume that $yz=\text{max}\{xy,yz,zx\}.\;$ As a consequence, $\displaystyle\frac{1}{3}\le yz\le 1.\;$ Define $y+z=2s\;$ and $yz=p;\;$ then, by the AM-GM inequality, $s\ge p\;$ and also $\displaystyle\frac{1}{\sqrt{3}}\le p\le 1.\;$ On the other hand, $\displaystyle x=\frac{1-xy}{x+y}=\frac{1-p^2}{2s}.\;$ Further $\sqrt{y}+\sqrt{z}=\sqrt{y+z+2\sqrt{yz}}=\sqrt{2s+2p}.\;$ For any fixed $\displaystyle p\in\left[\frac{1}{\sqrt{3}},1\right]\;$ we consider the function $f_p:\,[p,\infty )\rightarrow\mathbb{R},\;$ defined by $\displaystyle f_p(t)=\sqrt{\frac{1-p^2}{2t}}+\sqrt{2t+2p}.$

First off, $\displaystyle f'_{p}(t)=-\frac{\sqrt{1-p^2}}{(2t)^{3/2}}+\frac{1}{\sqrt{2t+2p}}.\;$ We'll prove that $f'_{p}(t)\ge 0.\;$ This is equivalent to showing that $8t^3\ge (1-p^2)(2t+2p),\;$ i.e., $4t^3-(1-p^2)t-(1-p^2)p\ge 0,\;$ for $t\ge p.$

Define function $g_p(t):\,[p,\infty)\rightarrow\mathbb{R},\;$ by $g_p(t)=4t^3-(1-p^2)t-(1-p^2)p.\;$ The only critical point of $g_p(t)\;$ in $[0,\infty)\;$ is $\displaystyle t=\sqrt{\frac{1-p^2}{12}},\;$ which is clearly less than $p,\;$ implying $g_p(t)\ge g_p(p)=2p(3p^2-1)\ge 0,\;$ for $t\ge p.\;$ This means that $f'_p(t)\ge 0,\;$ for $t\ge p\;$ such that $f_p(t)\;$ is strictly increasing for $t\ge p,\;$ so that $\displaystyle f_p(t)\ge f_p(p)=\sqrt{\frac{1-p^2}{2p}}+2\sqrt{p}\;$ for $t\ge p,\;$ $s,\;$ in particular.

Thus, suffice it to show that, for $\displaystyle p\in\left[\frac{1}{\sqrt{3}}\right],\;$ $\displaystyle\sqrt{\frac{1-p^2}{2p}}+2\sqrt{p}\ge 2.\;$ This is equivalent to $\displaystyle\sqrt{\frac{(1-p)(1+p)}{2p}}\ge\frac{2(1-p)}{1+\sqrt{p}}.\;$ Since $1-p\ge 0,$ we just need to prove $\displaystyle\sqrt{\frac{1+p}{2p}}\ge\frac{2\sqrt{1-p}}{1+\sqrt{p}}.\;$ Set $\sqrt{p}=u.\;$ Then $\displaystyle u\in\left[\frac{1}{\sqrt[4]{3}},1\right]\;$ and we'll show that $\displaystyle\frac{1+u^2}{2u^2}\ge\frac{4(1-u^2)}{(1+u)^2}\;$ which is $9u^4+2u^3-6u^2+2u\ge 0,\;$ or $(3u^2-1)^2+2u^3+2u\ge 0.\;$ The latter is obviously true for $\displaystyle u\in\left[\frac{1}{\sqrt[4]{3}},1\right].\;$ The proof is complete.

Solution 2

Let $\cot A = a^2,\ldots\;$ Then $a^2b^2+b^2c^2+c^2a^2=1\;$ which means $c=\sqrt{(1-a^2b^2)/(a^2+b^2)}\;$ and we have to show that $a+b+c\ge 2\;$ for $a,b,c\ge 0.\;$ If $s=a+b\ge 2,\;$ there is nothing to prove. Suppose $0

$f(a) = (1-a^2(s-a)^2)/(a^2+(s-a)^2)\;$ is even with repect to $a=s/2,\;$ $f'(0)=2/s^3\;$ is positive and $f'(a)\;$ has only three (possible) simple zeros: $s/2,\;$ $(s \pm\sqrt{-s^2 + 2 \sqrt{-4 + s^4}})/2.\;$ Thus the minimum of $f(a)\;$ for $0\le a\le s\lt 2\;$ is $\min(f(0),f(s/2)).\;$ Thus $a+b+c\ge s + \min(1,\sqrt(2-s^4/8))/s\;$ and it is easy to show that $\min(1,\sqrt{2-s^4/8})/s \ge 2-s.$


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