An All-Inclusive Inequality II

Problem

An  All-Inclusive Inequality II

Solution 1

$\displaystyle LHS\le\left(\sqrt{\sum_{cycl}m_a}\sqrt{\sum_{cycl}\frac{1}{h_a }}\right)\le (\sum_{cycl}m_a)\sqrt{\sum_{cycl}\frac{1}{h_a }}\sqrt{\sum_{cycl}\frac{1}{h_a }},$

for, $\ell_a\ge h_a,\;$ etc., implying $\displaystyle\sum\frac{1}{\ell_a}\le\sum\frac{1}{h_a}.\;$ Thus,

$\displaystyle\begin{align} LHS &\le\left(\sum_{cycl}m_a\right)\left(\sum_{cycl}\frac{1}{h_a}\right)\\ &=\left(\sum_{cycl}m_a\right)\left(\frac{s}{S}\right), \end{align}$

where $\displaystyle s=\frac{a+b+c}{2}\;$ is the semiperimeter, $S\;$ the area of $\Delta ABC.$

Using Bottema's inequality, $m_a+m_b+m_c\le 4R+r,\;$

$\displaystyle\begin{align} LHS&=\left(\sum_{cycl}m_a\right)\left(\frac{s}{S}\right)\\ &\le(4R+r)\left(\frac{s}{rs}\right)\\ &=\frac{4R+r}{r}. \end{align}$

Thus, suffice it to show that $\displaystyle\frac{4R+r}{r}\le\frac{9R}{2r},\;$ but this is equivalent to $8R+2r\le 9R,\;$ i.e., Euler's inequality $R\ge 2r.$

Solution 2

Let $AA'=m_a; OA'\perp BC; O \text{ - center of the circumcircle}\;(ABC).\;$

An  All-Inclusive Inequality II, illustration

In $\Delta ABC,$

$AA'\leq A'O+OA.$

It follows that

$\displaystyle\begin{align} m_a &\leq R\cos A+R= R(\cos A+1)=R\Bigr(2\cos^2 \frac{A}{2}-1+1\Bigr)\\ &=2R\cos^2 \frac{A}{2}. \end{align}$

Further,

$\displaystyle\begin{align} \frac{m_a}{h_a} &\leq \frac{2R\cos^2 \frac{A}{2}}{h_a}=\frac{2Rp(p-a)}{bc\cdot \frac{2S}{a}}=\frac{Rap(p-a)}{bcS}\\ &=\frac{Ra^2p(p-a)}{abcS}=\frac{Ra^2p(p-a)}{4RS^2}=\frac{a^2p(p-a)}{4S^2}=\frac{a^2p(p-a)}{4rpS}\\ &=\frac{a^2(p-a)}{4r^2p}. \end{align}$

Adding to that two analogous inequalities,

$\displaystyle\begin{align} \sum_{cycl}\frac{m_a}{h_a} &\leq \sum_{cycl}\frac{a^2(p-a)}{4r^2p}\\ &=\frac{1}{4r^2p}(p\sum a^2-\sum a^3). \end{align}$

We are going to use the following two identities:

$\displaystyle\sum_{cycl}a^2=2s^2-8Rr-2r^2,\\ \displaystyle\sum_{cycl}a^3=s(2s^2-12Rr-6r^2).$

With these,

$\displaystyle\begin{align} \sum \frac{m_a}{h_a}&\leq \frac{1}{4r^2p}\Bigr(p(2p^2-8Rr-2r^2)-p(2p^2-12Rr-6r^2)\Bigr)\\ &= \frac{1}{4r^2}\Bigr(2p^2-8Rr-2r^2-2p^2+12Rr+6r^2)\\ &=\frac{1}{4r^2}\Bigr(4Rr+4r^2\Bigr)=\frac{R}{r}+1. \end{align}$

For the record,

(1)

$\displaystyle\sum_{cycl}\frac{m_a}{h_a}\le\frac{R}{r}+1.$

From that and $\ell_a\ge h_a,\;$ etc., also

(2)

$\displaystyle\sum_{cycl}\frac{m_a}{\ell_a}\le\frac{R}{r}+1.$

By the Cauchy-Schwarz inequality, using (1),

(3)

$\displaystyle\begin{align} \Biggl(\sum \sqrt{\frac{m_a}{h_a}}\cdot 1\Biggl)^2 &\leq \Biggl(\sum \frac{m_a}{h_a}\Biggl)(1^2+1^2+1^2)=3\sum \frac{m_a}{h_a} &\leq 3\Bigr(\frac{R}{r}+1\Bigr). \end{align}$

Using 2, we similarly get

(4)

$\displaystyle \Biggl(\sum \sqrt{\frac{m_a}{\ell_a}}\cdot 1\Biggl)^2 \leq 3\Bigr(\frac{R}{r}+1\Bigr).$

The product of (3) and (4), along with the Euler inequality, $R\ge 2r\;$ yields the required inequality:

$\displaystyle\begin{align} \Biggl(\sum \sqrt{\frac{m_a}{l_a}}\Biggl)^2\Biggl(\sum \sqrt{\frac{m_a}{h_a}}\Biggl)^2 &\leq 9\Biggl(\frac{R+r}{r}\Biggl)^2\\ &\leq 9\Biggl(\frac{R+\frac{R}{2}}{r}\Biggl)^2 \end{align}$

and, finally,

$\displaystyle\begin{align} \Biggl(\sum \sqrt{\frac{m_a}{l_a}}\Biggl)\Biggl(\sum \sqrt{\frac{m_a}{h_a}}\Biggl)&\leq 3\frac{\frac{3R}{2}}{r}\\ &=\frac{9R}{2r}. \end{align}$

Solution 3

We shall use the following facts:

(1)

$\ell_a\ge h_a,\;\ell_b\ge h_b,\;\ell_c\ge h_c$

and

(2)

$\displaystyle\frac{R}{2r}\ge\frac{m_a}{h_a},\;\frac{R}{2r}\ge\frac{m_b}{h_b},\;\frac{R}{2r}\ge\frac{m_c}{h_c}.$

So we have

(A)

$\displaystyle\sqrt{\frac{m_a}{h_a}}+\sqrt{\frac{m_b}{h_b}}+\sqrt{\frac{m_c}{h_c}}\le 3\sqrt{\frac{R}{2r}}$

and, therefore, also

(B)

$\displaystyle\sqrt{\frac{m_a}{\ell_a}}+\sqrt{\frac{m_b}{\ell_b}}+\sqrt{\frac{m_c}{\ell_c}}\le 3\sqrt{\frac{R}{2r}}$

The product of (A) and (B) is exactly the required inequality.

Solution 4

We know that

$\ell_a\ge h_a,\;\ell_b\ge h_b,\;\ell_c\ge h_c,\\ m_a+m_b+m_c\le 4R+r\le\displaystyle\frac{9R}{2},\\ \displaystyle\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}=\frac{1}{r}.$

By the Cauchy-Schwarz inequality,

$\displaystyle\left(\sqrt{\frac{m_a}{h_a}}+\sqrt{\frac{m_b}{h_b}}+\sqrt{\frac{m_c}{h_c}}\right)^2\le (m_a+m_b+m_c)\left(\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}\right).$

It follows that

$\displaystyle\sqrt{\frac{m_a}{h_a}}+\sqrt{\frac{m_b}{h_b}}+\sqrt{\frac{m_c}{h_c}}\le \sqrt{\left(\frac{9R}{2}\right)}\sqrt{\left(\frac{1}{r}\right)}.$

If so, also

$\displaystyle\sqrt{\frac{m_a}{\ell_a}}+\sqrt{\frac{m_b}{\ell_b}}+\sqrt{\frac{m_c}{\ell_c}}\le \sqrt{\left(\frac{9R}{2}\right)}\sqrt{\left(\frac{1}{r}\right)}.$

The product of the two gives the required inequality.

Solution 5

We start with a series of equivalent inequalities:

$\displaystyle\frac{R}{2R}\ge\frac{m_a}{h_a},\;$ $\displaystyle\frac{R}{2R}\ge\frac{m_a}{\displaystyle\frac{2rs}{a}},\;$ $Rs\ge am_a,\;$ $\displaystyle\frac{abc}{4r}\ge am_a,\;$ $\displaystyle\frac{bc}{4r}\ge m_a,\;$ $b^2c^2\ge 4r^2(2b^2+2c^2-a^2),\;$ $b^2c^2\ge 4\displaystyle\frac{(s-a)(s-b)(s-c)}{s}(2b^2+2c^2-a^2),\;$ $b^2c^2s-4(s-a)(s-b)(s-c)(2b^2+2c^2-a^2)\ge 0.$

Let's make a substitution: $x=s-a,\;$ $y=s-b,\;$ $z=s-c,\;$ $a=y+z,\;$ $b=x+z,\;$ $c=x+y.\;$ In terms of $x,y,z\;$ the last inequality becomes

$(x+z)^2(x+y)^2(x+y+z)-4xyz[2(x+z)^2+2(x+y)^2-(y+z)^2]\ge 0.$

And further this can be reduced to

$y(x+y)^2(x-z)^2+z(x+z)^2(x-y)^2+x(x^2+xy+xz-3yz)^2\ge 0$

which is obviously true.

Acknowledgment

The problem has been by Dan Sitaru at the CutTheKnotMath facebook page; solutions added via comments and private communication. Solution 1 is by Soumava Chakraborty; Solution 2 is by Dan Sitaru; Solutions 3, 4 and 5 are by Kevin Soto Palacios. The problem came from Dan's book Math Power and has been published at the Romanian Mathematical Magazine.

 

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