Walker's inequality.

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Walker's inequality.

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Source: Some proofs.

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#1 h Apr 2, 2010, 5:10 PM • 2 Y

Y by Adventure10, Mango247

Prove that in any acute or right triangle exists the Walker's inequality

$a^2 + b^2 + c^2\ \ge\ 4(R + r)^2$ , i.e. an equivalent form $\boxed {\ s^2\ \ge\ 2R^2 + 8Rr + 3r^2\ }$

with equality if and only if $\triangle ABC$ is equilateral or right isosceles.

Remark. I have two own (?) elementary proofs of it. Maybe someone knows a link with its another elementary proofs. Thank you.

This post has been edited 1 time. Last edited by Virgil Nicula, Apr 2, 2010, 6:28 PM

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#2 h Apr 2, 2010, 5:54 PM • 2 Y

Y by Adventure10, Mango247

Virgil Nicula wrote:

Prove that in any acute or right triangle exists the Walker's inequality

$a^2 + b^2 + c^2\ \ge\ 4(R + r)^2$ , i.e. an equivalent form $\boxed {\ s^2\ \ge\ 2R^2 + 8Rr + 3r^2\ }$

with equality if and only if $\triangle ABC$ is equilateral or right isosceles.

Remark. I have two own (?) elementary proofs of it. Maybe someone knows a link with its another proofs. Thank you.

an other equivalent form:

$\boxed{a^2 + b^2 + c^2 \geq (AH + BH + CH)^2}$

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#3 h Apr 2, 2010, 6:14 PM • 1 Y

Y by Adventure10

Virgil Nicula wrote:

Prove that in any acute or right triangle exists the Walker's inequality

$a^2 + b^2 + c^2\ \ge\ 4(R + r)^2$ , i.e. an equivalent form $\boxed {\ s^2\ \ge\ 2R^2 + 8Rr + 3r^2\ }$

with equality if and only if $\triangle ABC$ is equilateral or right isosceles.

Remark. I have two own (?) elementary proofs of it. Maybe someone knows a link with its another proofs. Thank you.

Since

$HA = 2R \cos A, HB = 2R \cos B, HC = 2R \cos C$

and Walker's are equivalent to

$\sin^2 A + \sin^2 B + \sin^2 C \geq (\cos A + \cos B + \cos C )^2$

Making the pedal triangle trasformation

$A = \frac{\pi - A'}{2}, B = \frac{\pi - B'}{2}, C = \frac{\pi - C'}{2}$

we have the seguent equivalent

$\cos^2 \frac{A'}{2} + \cos^2 \frac{B'}{2} + \cos^2 \frac{C'}{2} \geq ( \sin \frac{A'}{2} + \frac{B'}{2} + \frac{C'}{2})^2$ $\Leftrightarrow$ BOTTEMA 2.56. Q.V.D.

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#4 h Apr 3, 2010, 4:47 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Virgil Nicula wrote:

Prove that in any acute or right triangle exists the Walker's inequality $a^2 + b^2 + c^2\ \ge\ 4(R + r)^2$ .

Method 1 (own). $\frac {a^2 + b^2 + c^2}{8RS} = \sum\frac {\cos^2A}{a\cdot\cos A}\ \stackrel{(C.B.S.)}{\ge}\ \frac {\left(\sum\cos A\right)^2}{\sum a\cdot\cos A} =$ $\frac {(R + r)^2}{2RS}\ \implies\ a^2 + b^2 + c^2\ \ge\ 4(R + r)^2$ .

Remark. $HA = 2\cdot\delta_{BC}(O)$ and $\sum\delta_{BC}(O) = R + r$ (Euler's relation in an acute triangle) $\implies\ \sum HA = 2(R + r)$

$\implies\ \boxed {\ \sum a^2\ge \left(\sum HA\right)^2\ }$ . I denote $\delta_{BC}(O)$ - the distance of the circumcenter $O$ of $\triangle ABC$ to the sideline $BC$ .

Method 2 (Cezar Lupu). $2 = \sum\frac {HA}{h_a} =$ $\sum\frac {HA^2}{h_a\cdot HA}\ \stackrel{(C.B.S.)}{\ge}\ \frac {\left(\sum HA\right)^2}{\sum h_a\cdot HA} =$ $\frac {\left(\sum HA\right)^2}{\frac 12\cdot\sum a^2}\ \implies\ \sum a^2\ge\ \left(\sum HA\right)^2$ .

I used well-known relations $bc = 2Rh_a$ a.s.o. and $b^2 + c^2 - a^2 = 2bc\cdot\cos A$ a.s.o. See here the Cezar Lupu's article.

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#5 h Apr 7, 2010, 6:01 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

(continuation)

Remark. In an acute triangle $ABC$ there are the inequalities $\|\begin{array}{c} \sum \tan A\ \ge\ \frac sr\ \ge\ 3\sqrt 3 \\ \\ \sum\cot A\ \ge\ \frac {(R + r)^2}{S}\ \ge\ \sqrt 3 \\ \\ \sum\frac {1}{\sin 2A}\ \ge\ \frac {9R^2}{2S}\ \ge\ \frac {s^2 + (R + r)^2}{2S}\ \ge\ \frac {2(4R + r)}{s}\ \ge\ 2\sqrt 3\end{array}\|$ .

Can obtain the second inequality with the well-known metrical relation $4S = (b^2 + c^2 - a^2)\cdot\tan A$ .

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Mateescu Constantin

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Mateescu Constantin

#6 h Jul 13, 2010, 1:30 PM • 2 Y

Y by Adventure10, Mango247

My own proof for Walker's inequality.

Firstly, I will prove the following lemma :

Quote:

Lemma. In any triangle $ABC$ with incenter $I$ the following inequality holds : $\boxed{\ AI+BI+CI\ \le\ 2(R+r)\ }$ .

Proof. Show easily that $AI=\frac{bc}s\cdot\cos\frac A2=\frac 1s\cdot\sqrt{bc}\cdot\sqrt {s(s-a)}$ a.s.o. Thus, we have :

$\left(\sum AI\right)^2=\frac{1}{s^2}\cdot\left(\sum\sqrt {bc}\cdot\sqrt{s(s-a)}\right)^2\ \stackrel{C.B.S.}{\le}\ \frac 1{s^2}\cdot (ab+bc+ca)\cdot\sum s(s-a)$ $=ab+bc+ca\ \le\ 4(R+r)^2$

The last inequality is due to Gerretsen i.e. $\boxed{\ s^2\ \le\ 4R^2+4Rr+3r^2\ }$ . Therefore, we have shown that : $\boxed{\ AI+BI+CI\ \le\ 2(R+r)\ }$ .

Now, for an acute triangle $ABC$ denote $H$ its orthocenter and let $AA_1$ , $BB_1$ , $CC_1$ be its altitudes . Thus, $H$ is the incenter of the orthic triangle $A_1B_1C_1$ .

Applying the upper lemma for $\triangle\ A_1B_1C_1$ we have : $HA_1+HB_1+HC_1\ \le\ 2(R_0+r_0)$ , where $R_0$ and $r_0$ are the circumradius and inradius of $\triangle A_1B_1C_1$ .

Taking into account that $\boxed{\ R_0=\frac R2\ }$ and $\boxed{\ r_0=2R\prod\cos A\ }$ , the last inequality becomes : $\sum\ (h_a-AH)\ \le\ 2\left(\frac R2+2R\prod\cos A\right)\ \iff$

$\iff\ h_a+h_b+h_c\ \le\ R+4R\prod\cos A+AH+BH+CH\ \iff\color{white}{.}$ $\frac{ab+bc+ca}{2R}\ \le\ R+4R\cdot\frac{s^2-(2R+r)^2}{4R^2}+2(R+r)\ \iff$

$\iff\ s^2+r^2+4Rr\ \le\ 2R^2+2s^2-2(2R+r)^2+4R^2+4Rr\color{white}{.}$ $\iff\ \boxed{\ 2R^2+8Rr+3r^2\ \le\ s^2\ }\ \iff\color{white}{.}$ Walker's inequality .

Remark. In the above proof I used the well-known relations $\boxed{\ \prod\cos A=\frac{s^2-(2R+r)^2}{4R^2}\ }$ and $\boxed{\ AH+BH+CH=2(R+r)\ }$ (in an acute triangle) .

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#7 h Jul 14, 2010, 1:02 AM • 2 Y

Y by Adventure10, Mango247

Thank you, very nice your proofs !

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Potla

#8 h Aug 28, 2010, 4:10 AM • 6 Y

Y by Mateescu Constantin, Adventure10, Mango247, and 3 other users

Ligouras wrote:

Virgil Nicula wrote:

Prove that in any acute or right triangle exists the Walker's inequality

$a^2 + b^2 + c^2\ \ge\ 4(R + r)^2$ , i.e. an equivalent form $\boxed {\ s^2\ \ge\ 2R^2 + 8Rr + 3r^2\ }$

with equality if and only if $\triangle ABC$ is equilateral or right isosceles.

Remark. I have two own (?) elementary proofs of it. Maybe someone knows a link with its another proofs. Thank you.

an other equivalent form:

$\boxed{a^2 + b^2 + c^2 \geq (AH + BH + CH)^2}$

We rewrite this into
$\sin^2A+\sin^2B+\sin^2C\geq (\cos A+\cos B+\cos C)^2;$
But we know the identities: $\displaystyle \left\|\begin{array}{ccc}{2\cos A\sin B\sin C=\sin^2B+\sin^2C-\sin^2A}\\{\displaystyle\frac{\cos A}{\sin B\sin C}+\frac{\cos B}{\sin C\sin A}+\frac{\cos C}{\sin A\sin B}=2}\end{array}\right\|;$
And so we apply the Cauchy-Schwarz inequality as follows:
$\begin{aligned}\left(\cos A+\cos B+\cos C\right)^2&=\left(\sum_{cyc}\sqrt{\cos A\sin B\sin C}\sqrt{\frac{\cos A}{\sin B\sin C}}\right)^2\\& \leq \left(\sum_{cyc}\cos A\sin B\sin C\right)\cdot \sum_{cyc}\frac{\cos A}{\sin B\sin C}\\&=\sum_{cyc}\left(\sin^2B+\sin^2C-\sin^2A\right);\end{aligned}$
And we are done. $\Box$

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