The Weitzenböck inequality: a^2 + b^2 + c^2 >= 4 sqrt(3) S

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Iris Aliaj

165 posts

Iris Aliaj

#1 h Jul 21, 2004, 5:59 PM • 5 Y

Y by Adventure10, megarnie, jhu08, Mango247, and 1 other user

Let $a$ , $b$ , $c$ be the sides of a triangle, and $S$ its area. Prove:
$a^{2} + b^{2} + c^{2}\geq 4S \sqrt {3}$
In what case does equality hold?

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fuzzylogic

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fuzzylogic

#2 h Jul 21, 2004, 6:20 PM • 10 Y

Y by machffud.thvdrk18, Bottp, SardorNarpay, Adventure10, jhu08, Seungjun_Lee, Mango247, ehuseyinyigit, aidan0626, and 1 other user

$a^2+b^2+c^2-4\sqrt{3}S = a^2+b^2 + (a^2+b^2-2ab\cos{C})-2\sqrt{3}ab\sin{C}$ $= 2(a-b)^2 + 4ab(1-\cos(C+\pi/3)) \geq 0$

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darij grinberg

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darij grinberg

#3 h Jul 21, 2004, 7:38 PM • 7 Y

Y by tsotnedadiani, Bottp, Adventure10, jhu08, Mango247, and 2 other users

If a, b, c are the sidelengths of a triangle and S is its area, then

$4 \sqrt3 S \leq a^2 + b^2 + c^2$ .

Indeed there are lots of proofs for this inequality. Here is my favorite proof:

Let X be such a point that the triangle BCX is equilateral, and the points A and X lie in the same halfplane with respect to the line BC. Then, after the Cosine Law in triangle ABX, we have

$AX^{2} = AB^{2} + BX^{2} - 2\cdot AB\cdot BX\cdot \cos \measuredangle ABX$ .

But BX = BC and < XBC = 60°, since the triangle BCX is equilateral; hence < ABX = < ABC - < XBC = < ABC - 60° = B - 60°, and we get

$AX^{2}=AB^{2}+BC^{2}-2\cdot AB\cdot BC\cdot \cos \left( B-60^{\circ }\right)$
$=c^{2}+a^{2}-2ca\cos \left( B-60^{\circ }\right)$
$=c^{2}+a^{2}-2ca\left( \cos B\cos 60^{\circ }+\sin B\sin 60^{\circ }\right)$
$=c^{2}+a^{2}-2ca\left( \frac{1}{2}\cos B+\frac{\sqrt{3}}{2}\sin B\right)$
$=c^{2}+a^{2}-ca\cos B-\sqrt{3}ca\sin B$
$=c^{2}+a^{2}-\frac{c^{2}+a^{2}-b^{2}}{2}-\sqrt{3}ca\sin B\text{\ \ \ \ \ \ \ \ \ \ (Cosine Law)}$
$=\frac{a^{2}+b^{2}+c^{2}}{2}-\sqrt{3}ca\sin B$
$=\frac{a^{2}+b^{2}+c^{2}}{2}-\sqrt{3}\cdot 2S\text{\ \ \ \ \ \ \ \ \ \ (since }S=\frac{1}{2}ca\sin B\text{)}$ ;

thus, $\displaystyle \frac {a^{2} + b^{2} + c^{2}}{2}\geq \sqrt {3}\cdot 2S$ , so that $a^{2} + b^{2} + c^{2}\geq 4\sqrt {3}S$ , what was to be proven. Equality occurs if and only if $AX^{2} = 0$ , i. e. if the points A and X coincide, i. e. if the triangle ABC is equilateral.

Darij

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blahblahblah

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blahblahblah

#4 h Aug 9, 2004, 6:34 AM • 4 Y

Y by Adventure10, jhu08, Mango247, and 1 other user

fuzzylogic wrote:

$a^2+b^2+c^2-4\sqrt{3}S = a^2+b^2 + (a^2+b^2-2ab\cos{C})-2\sqrt{3}ab\sin{C}$ $= 2(a-b)^2 + 4ab(1-\cos(C+\pi/3)) \geq 0$

Like darij, I have seen many nice proofs of this inequality, but that is just beautiful!

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Iris Aliaj

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Iris Aliaj

#5 h Aug 21, 2004, 11:11 AM • 5 Y

Y by jhu08, Adventure10, Vladimir_Djurica, Mango247, and 1 other user

Weitzenböck's inequality: Prove that if a,b,c are the side lengths of a triangle and Sa it area then: 4S√3 ≤ a 2 + b 2 + c 2

To prove the inequality of Weitzenbock, we should first prove an aiding inequality.
xy+xz+yz ≥ √3xyz(x+y+z)
(xy+xz+yz) 2 ≥ 3xyz(x+y+z)
(xy+xz+yz) 2 = (xy) 2 +(xz) 2 +(yz) 2 +2xyz(x+y+z) ≥ ⅓(xy+xz+yz) 2 +2xyz(x+y+z)
Therefore
⅔(xy+xz+yz) 2 ≥ 2xyz(x+y+z)
(xy+xz+yz)2 ≥ 3xyz(x+y+z) We now have the required inequality.
Going back to the starting inequality.
S√3 = √3p(p-a)(p-b)(p-c) = √3(p-a+p-b+p-c)(p-a)(p-b)(p-c)
Apply the aiding inequality and get:
S :sqrt: 3<=(p-a)(p-b)+(p-b)(p-c)+(p-a)(p-c)
= ab+bc+ac-p 2
Therefore
4S√3 ≤ 4ab+4bc+4ac-(a+b+c) 2
= a 2 +b 2 +c 2 -(a-b) 2 -(b-c) 2 -(a-c) 2
4s√3 ≤ a2 + b2 + c2
Therefore, Weitzenbock�s inequality is now proved.

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nickolas

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nickolas

#6 h Sep 14, 2004, 9:16 AM • 5 Y

Y by jhu08, Adventure10, Hexagrammum16, Vladimir_Djurica, Mango247

1)Aply am-gm to $s-a , s-b , s-c$ and find that $s\geq3\sqrt{3}r$
2). $a^2+b^2+c^2\geq\frac{1}{3}(a+b+c)^2=\frac{4}{3}s^2\geq4\sqrt{3}sr=4\sqrt{3}A$

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Michael Niland

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Michael Niland

#7 h Jan 22, 2005, 6:45 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Consider the squares of each side of inequality we want to prove.Squaring on the left we get
a^4 + b^4 +c^4 +2(a^2)(b^2 ) +2(b^2)(c^2) +2(c^2)(a^2) and

3.16(S^2) =3 [2(a^2)(b^2) +2(b^2)(c^2) +2(c^2)(a^2) -(a^4) -(b^4) -(c^4)] on the right.
The difference between these two expressions is easily seen to be positive.

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Vasc

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Vasc

#8 h Jan 22, 2005, 9:49 AM • 3 Y

Y by jhu08, Adventure10, Mango247

This is a very weak inequality because
$a^2+b^2+c^2 \ge (a+b+c)^2/3 \ge ab+bc+ca \ge 3(abc)^{2/3} \ge 4S\sqrt{3}$ .

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blahblahblah

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blahblahblah

#9 h May 29, 2005, 9:45 PM • 3 Y

Y by Adventure10, jhu08, Mango247

It is not necessary to post meaningless things like 'this is a weak inequality' to threads that have been dead for nearly 5 months.

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Cezar Lupu

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Cezar Lupu

#10 h Oct 8, 2005, 5:04 PM • 3 Y

Y by Adventure10, jhu08, Mango247

Come on, this inequality has benn posted many times before.
The following inequalities hold:

$a^{2}+b^{2}+c^{2}\geq ab+bc+ca\geq a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}\geq 18Rr\geq 4S\sqrt{3}$

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Cezar Lupu

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Cezar Lupu

#11 h Oct 8, 2005, 11:08 PM • 4 Y

Y by Adventure10, Adventure10, jhu08, Mango247

Although here is a stronger one due to Hadwiger-Finsler, 1937 which states that

$a^{2}+b^{2}+c^{2}\geq 4S\sqrt{3}+(a-b)^{2}+(b-c)^{2}+(c-a)^{2}.$

[Moderator edit: See http://www.mathlinks.ro/Forum/viewtopic.php?t=31598 for proofs of this inequality.]

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Virgil Nicula

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Virgil Nicula

#12 h Oct 11, 2005, 9:31 PM • 2 Y

Y by Adventure10, jhu08

Caesar, how show you the inequality $\sum a\sqrt{bc}\ge 18Rr \ ?$

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Cezar Lupu

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Cezar Lupu

#13 h Oct 11, 2005, 9:38 PM • 3 Y

Y by Adventure10, jhu08, Mango247

Nice and easy

. Because $2Rr=\frac{abc}{a+b+c}$ the inequality reduces to the following:

$\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\geq\frac{9}{a+b+c}$ .
By AM-GM we have, $\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\geq\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}$ and from here it is clear how to proceed.

P.S. I have found some nice stronger inequalities connected to these ones.

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perfect_radio

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perfect_radio

#14 h Nov 5, 2005, 11:56 PM • 2 Y

Y by Adventure10, jhu08

Vasc wrote:

This is a very weak inequality because
$\ldots \geq 3(abc)^{2/3} \geq 4S\sqrt{3}$ .

How do we prove the last inequality ?

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Cezar Lupu

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Cezar Lupu

#15 h Nov 6, 2005, 6:38 PM • 3 Y

Y by Adventure10, jhu08, Mango247

By applying AM-GM inequality.

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mihaig

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mihaig

#58 h Sep 20, 2021, 5:54 AM

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Math-wiz wrote:

mihaig wrote:

In $\Delta ABC$ with area $S,$ , prove
$a^2+b^2+c^2\geq4\sqrt3\cdot S+2\left(a-b\right)^2.$ When do we have equality?

Roberto Tauraso, 2017.

We need to prove $c^2-a^2-b^2+4ab\geq 4\sqrt{3}S$ Substituting $c^2-a^2-b^2=-2ab\cos C$ and $ab=\frac{2S}{\sin C}$ , we need to prove $2\csc C-\cot C\geq\sqrt{3}$
Let $\csc C=x$
We need to prove $x-\sqrt{x^2-1}\geq\sqrt{3}$ which is equivalent to $(\sqrt{3}x-2)^2\geq 0$ Equality when $\angle C=\frac{\pi}{3},\frac{2\pi}{3}$

Bravo

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ZETA_in_olympiad

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ZETA_in_olympiad

#59 h Apr 2, 2022, 5:20 AM

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By Heron's formula and AM-GM,
$16S^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) \leq (a+b+c)(\frac{a+b+c}{3})^3$
$\implies 4S\leq \frac{(a+b+c)^2}{3\sqrt{3}} =\sqrt{3}(\frac{a+b+c}{3})^2 \leq \sqrt{3}\frac{a^2+b^2+c^2}{3}$
$\implies a^2+b^2+c^2 \geq 4S \sqrt{3}.$
Equality when $a=b=c.$
[Note: I thought the area was $A$ !]

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sqing

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sqing

#60 h Apr 2, 2022, 6:33 AM • 1 Y

Y by ZETA_in_olympiad

Let $a$ , $b$ , $c$ be the sides of a triangle, and $S$ its area. Prove:
$a^2+ b^2+ c^2\geq 4S \sqrt {3}+\frac{( b^2- c^2)^2}{2a^2}$ SXTB,(3)2022,Q2651

Iris Aliaj wrote:

Let $a$ , $b$ , $c$ be the sides of a triangle, and $S$ its area. Prove:
$a^{2} + b^{2} + c^{2}\geq 4S \sqrt {3}$ In what case does equality hold?

North Macedonian 1998

sqing wrote:

stronger than Weitzenbock, a present for nguyentrang: $a^2+b^2+c^2\ge 4\sqrt{4-sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}}S.$

In triangle $ABC,$ show that $\tan{\frac{A}{2}}+\tan{\frac{B}{2}}+\tan{\frac{C}{2}} \ge \sqrt{4-8\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}}.$

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mihaig

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mihaig

#61 h Apr 2, 2022, 9:31 AM

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sqing wrote:

Iris Aliaj wrote:

Let $a$ , $b$ , $c$ be the sides of a triangle, and $S$ its area. Prove:
$a^{2} + b^{2} + c^{2}\geq 4S \sqrt {3}$ In what case does equality hold?

North Macedonian 1998

Ionescu, early 1900

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#62 h May 29, 2022, 8:07 AM

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With using Heron's formula we need to prove $(a^2+b^2+c^2)^2 \ge 48p(p-a)(p-b)(p-c)$ or $(a^2+b^2+c^2)^2 \ge 3(a+b+c)(b+c-a)(c+a-b)(a+b-c)$ . Note that by Ravi and simple AM-GM we have $(b+c-a)(c+a-b)(a+b-c) \le abc$ so we need to prove $(a^2+b^2+c^2)^2 \ge 3(a+b+c)abc$ or $(ab+bc+ca)^2 \ge 3(a+b+c)abc$ which is true.

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mihaig

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mihaig

#63 h May 30, 2022, 2:27 PM • 3 Y

Y by Mango247, Mango247, Mango247

Interesting

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lifeismathematics

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lifeismathematics

#64 h Nov 21, 2022, 12:16 PM • 1 Y

Y by Mango247

indeed a good one

This post has been edited 2 times. Last edited by lifeismathematics, Nov 21, 2022, 12:21 PM

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prMoLeGend42

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prMoLeGend42

#65 h Jan 16, 2023, 4:38 AM

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Let $s$ be the semi-perimeter and $\triangle$ be the area of the triangle respectively .

Lemma: $s^2\geq 3\sqrt{3}\triangle$

Proof : By AM-GM , $\frac{(s-a)+(s-b)+(s-c)}{3}\geq \sqrt[3]{(s-a)(s-b)(s-c)}\implies s\geq 3 \sqrt[3]{\frac{{\triangle}^2}{s}} \implies s^4\geq 27{\triangle}^2$ $\implies \boxed{s^2\geq 3 \sqrt{3} \triangle}$

Now , by Titu's Lemma , we have

$a^2+b^2+c^2\geq \frac{(a+b+c)^2}{3}=\frac{4s^2}{3} \geq 4\sqrt{3}\triangle$

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huashiliao2020

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huashiliao2020

#66 h Jun 10, 2023, 4:31 AM

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$\blacksquare$

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Matematika_Sejati

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Matematika_Sejati

#67 h Jun 10, 2023, 6:40 AM

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There's ma solution

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Aoxz

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Aoxz

#68 h Jun 10, 2023, 8:51 AM • 1 Y

Y by Yunis019

$a^{2} + b^{2} + c^{2}\geq 4S \sqrt {3}$ Let $a=x+y$ , $b=y+z$ , $c=z+x$
INEQUALITY becomes $2(x^2+y^2+z^2+xy+yz+zx) \geq 4\sqrt{3xyz(x+y+z)}$ well known lemma $xy+yz+zx \geq \sqrt{3xyz(x+y+z)}$
well known lemma $x^2+y^2+z^2 \geq xy+yz+zx$
and rest is easy
Aoxz

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mathmax12

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mathmax12

#69 h Jul 18, 2023, 11:57 PM

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$a^2+b^2+c^2-4\sqrt{3}S=2(a^2+b^2)-2ab\cos(C)-2\sqrt{3}ab\sin(C)=2(a-b)^2+4ab(1-\cos(C+\pi/3)) \ge 0$ , equality holds when $\cos(C+\pi/3)=0$ , which holds when the triangle is equalateral.

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iMqther_

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iMqther_

#71 h Aug 22, 2024, 6:16 PM

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A solution using only basic geometry and inequality concepts:

Now notice that $a^2+b^2+c^2\geq4S \sqrt{3} \iff a^2+b^2+c^2\geq \sqrt{48S^2}$ .

By the Heron Formula we have $S=\sqrt{p(p-a)(p-b)(p-c)}$ where $p=\frac{a+b+c}{2}$ .

$S^2=p(p-a)(p-b)(p-c)$
$S^2=(\frac{a+b+c}{2})(\frac{b+c-a}{2})(\frac{a+c-b}{2})(\frac{a+b-c}{2})$
$16S^2=[(b+c+a)(b+c-a)(a+c-b)(a-(c-b)]$
$48S^2=3[(b+c)^2-a^2)(a^2-(b-c)^2]$
$\sqrt{48S^2}=\sqrt{3[(b+c)^2-a^2)(a^2-(b-c)^2]}$

With this we have that $a^2+b^2+c^2\geq\sqrt{48S^2} \implies a^2+b^2+c^2\geq\sqrt{3[(b+c)^2-a^2)(a^2-(b-c)^2]}$

$a^2+b^2+c^2\geq\sqrt{3[(b+c)^2-a^2)(a^2-(b-c)^2]} \iff$
$(a^2+b^2+c^2)^2\geq3[(b+c)^2-a^2)(a^2-(b-c)^2] \iff$
$(a^2+b^2+c^2)^2\geq3[(b+c)(b-c)]^2+[(b+c)a]^2-a^4+[(a(b-c)^2)] \iff$
$(a^2+b^2+c^2)^2\geq3[(b^2-c^2)^2+(ab+ac)^2-a^4+(ab-ac)^2] \iff$
$(a^2+b^2+c^2)^2\geq3(b^4+c^4-2b^2c^2+a^2b^2+a^2c^2+2a^2bc-a^4+a^2b^2+a^2c^2-2a^2bc)\iff$
$(a^2+b^2+c^2)^2\geq3b^4+3c^4-6b^2c^2+6b^2c^2+6a^2c^2-3a^4 \iff$
$2b^4+2c^4-8b^2c^2+4a^2b^2+4a^2c^2-4a^4\le0 \iff$
$b^4+c^4-2b^2c^2-2b^2c^2+2a^2c^2-2a^4\le0 \iff$
$(b^2-c^2)^2+2[(-a^2(a^2-b^2)+c^2(a^2-b^2)]\le0 \iff$
$(b^2-c^2)^2+2(a^2+b^2)(c^2-a^2)\le0.$

Note that this inequality is symmetric in $a,b,c$ so if we let $a$ be the greater side of the triangle we'll have that $c^2-a^2<0 \implies 2(a^2-b^2)(c^2-a^2)\le0$ but $(b^2-c^2)^2\geq0$ .

Therefore we have that $(b^2-c^2)^2+2(a^2+b^2)(c^2-a^2)\le0. \iff$
$b^2-c^2\le a^2-b^2 \iff a^2+c^2\geq 2b^2$ and this is true because since they're sides of a triangle then $a+c>b$ $\implies a^2+c^2>b^2$ so $a^2+c^2\geq 2b^2$ and this finishes the first part.

Equality holds iff $(b^2-c^2)^2+2(a^2+b^2)(c^2-a^2)=0$ in other words when $b^2=c^2, a^2=b^2$ or $b^2=c^2, c^2=a^2$ but in both cases we get $a=b=c$ , an equilateral triangle. $\square$ .

This post has been edited 2 times. Last edited by iMqther_, Aug 22, 2024, 6:20 PM
Reason: Typo

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ihatemath123

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ihatemath123

#72 h Aug 22, 2024, 9:22 PM • 1 Y

Y by OronSH

If the triangle is not scalene, assume WLOG that $AB \neq AC$ . We can decrease the LHS while fixing the RHS by replacing $A$ with the point $A'$ s.t. $A'B=A'C$ and $\overline{AA'} \parallel \overline{BC}$ . Now, it suffices to check the inequality when the triangle is equilateral, in which case it is obviously true.

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MathProGenius

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MathProGenius

#73 h Mar 11, 2025, 6:20 AM

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Before start, let's take a square of both sides:
(a²+b²+c²)²≥48S²
Using Heron formula for the area of triangle:
(a²+b²+c²)²≥48(a+b-c)(c+a-b)(c+b-a)(a+b+c)/16
Simplifying the expression:
a⁴+b⁴+c⁴≥a²b²+b²c²+a²c²
To prove this equation,we'll use AM-GM inequality:
½(a⁴+b⁴)≥a²b² (1)
½(a⁴+c⁴)≥a²c² (2)
½(c⁴+b⁴)≥c²b² (3)

(1)+(2)+(3): a⁴+b⁴+c⁴≥a²b²+b²c²+a²c²

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