Two Refinements of the Ionescu-Weitzenböck Inequality

Problem

D. M. Batinetu-Giurgiu and N. Stanciu
REVISTA DE MATEMATICA MEHEDINTEANA
Vol. 16, 2016, pp 14-16

Two Refinements of Ionescu-Weitzenbööck Inequality

Proof

WLOG, assume $am_a=\max\{am_a,bm_b,cm_c\}.\;$ Then the first inequality is equivalent to:

$\begin{align}(a^2+b^2+c^2)^2&\ge 12a^2m_a^2=(3a^2)(4m_a^2)\\ &=3a^2(2b^2+2c^2-a^2)\\ &=6a^2b^2+6a^2c^2-3a^4. \end{align}$

which is equivalent to

$a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2\ge 6a^2b^2+6a^2c^2-3a^4,$

or,

$4a^4+b^4+c^4-4a^2b^2-4a^2c^2+2b^2c^2\ge 0.$

But this is the same as

$(b+2+c^2-2a^2)^2\ge 0$

which is obviously true.

Now, for the second inequality. WLOG, $a\le b\le c.\;$ If so $\ell_a\ge\ell_b\ge\ell_c\;$ and $h_a\ge h_b\ge h_c.$

So by Chebyshev's inequality,

(1)

$\displaystyle\frac{\ell_a}{h_a}+\frac{\ell_b}{h_b}+\frac{\ell_c}{h_c}\le\frac{1}{3}(\ell_a+\ell_b+\ell_c)\left(\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}\right).$

But

$\displaystyle\begin{align} \frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}&=\frac{a}{ah_a}+\frac{b}{bh_b}+\frac{c}{ch_c}\\ &=\frac{a+b+c}{2S}\\ &=\frac{s}{S}, \end{align}$

where $S\;$ is the area, $s\;$ the semiperimeter of $\Delta ABC.$ Thus

(2)

$\displaystyle \frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}=\frac{s}{S}.$

We also have

$\displaystyle\begin{align} \ell_a&=\frac{2bc}{b+c}\cos\frac{A}{2}=\frac{2bc}{b+c}\sqrt{\frac{s(s-a)}{bc}}\\ &=\frac{2\sqrt{bc}}{b+c}\sqrt{s(s-a)}\\ &\le\frac{2\sqrt{bc}}{2\sqrt{bc}}\sqrt{s(s-a)}\\ &=\sqrt{s(s-a)}. \end{align}$

And we see that

(3)

$\ell_a\le\sqrt{s(s-a)}.$

and similarly for the other two. Therefore, we deduce that

(4)

$\ell_a+\ell_b+\ell_c\le\sqrt{s}(\sqrt{s-a}+\sqrt{s-b}+\sqrt{s-c}).$

Applying Jensen's inequality to the convex function $f(x)=x^2\;$ we obtain

$\displaystyle x^2+y^2+z^2\ge 3\left(\frac{x+y+z}{3}\right)^2=\frac{(x+y+z)^2}{3}$

which is equivalent to

(5)

$x+y+z\le \sqrt{3(x^2+y^2+z^2)}.$

We take here $x=\sqrt{s-a},\;$ $y=\sqrt{s-b},\;$ $ z=\sqrt{s-c}\;$ to obtain

$\displaystyle (s-a)+(s-b)+(s-c)\ge\left(\frac{\sqrt{s-a}+\sqrt{s-b}+\sqrt{s-c}}{3}\right)^2$

which reduces to

$\displaystyle 3s\ge\left(\frac{\sqrt{s-a}+\sqrt{s-b}+\sqrt{s-c}}{3}\right)^2$

and, subsequently, to

(6)

$\displaystyle\sqrt{s}\cdot\sqrt{3}\ge\frac{\sqrt{s-a}+\sqrt{s-b}+\sqrt{s-c}}{3}.$

From (4) and (6),

(7)

$\ell_a+\ell_b+\ell_c\le\sqrt{s}\sqrt{s}\sqrt{3}=s\sqrt{3}.$

Hence, by (1), (2), and (7),

$\displaystyle\begin{align} \frac{\ell_a}{h_a}+\frac{\ell_b}{h_b}+\frac{\ell_c}{h_c} &\le\frac{1}{3}\cdot s\sqrt{3}\cdot\frac{s}{S}=\frac{s^2}{S\sqrt{3}}\\ &=\frac{(a+b+c)^2}{4S\sqrt{3}}\\ &\le\frac{a^2+b^2+c^2}{4S}\sqrt{3} \end{align}$

and we are done.

Remark

The two inequalities (a) and (b) are in fact refinements of the Ionescu-Weitzenböck inequality. Indeed:

  • If in (a) we take into account that $m_a\ge h_a,\;$ we obtain

    $\begin{align} a^2+b^2+c^2 &\ge 2\sqrt{3}\cdot a\cdot m_a \ge 2\sqrt{3}\cdot a\cdot h_a\\ &=2\sqrt{3}\cdot 2S=4S\sqrt{3}. \end{align}$

  • If in (b) we take into account of $\ell_a\ge h_a,\;$ etc., (b) yields

    $\displaystyle 3\le\frac{\ell_a}{h_a}+\frac{\ell_b}{h_b}+\frac{\ell_c}{h_c}\le\frac{a^2+b^2+c^2}{4S}\sqrt{3}.$

    which is the Ionescu-Weitzenböck inequality.

Acknowledgment

I am indebted to Dan Sitaru for sending me a copy of the REVISTA DE MATEMATICA MEHEDINTEANA magazine.

 

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