$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$

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

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!

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.

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 $

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.

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

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

Come on, this inequality has benn posted many times before. The following inequalities hold: $a^{2}+b^{2}+c^{2}\geq ab+bc+ca\geq a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}\geq 18Rr\geq 4S\sqrt{3}$

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.]

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

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.

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 ?

By applying AM-GM inequality.

Cezar, please give a complete proof.

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

cezar lupu wrote: By applying AM-GM inequality. Could you develop a little on this?

perfect_radio wrote: 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 ? Try popoviciu on f(x) = ln x You can get $\prod (\frac{x+y}{2})^2 \ge xyz(\frac{x+y+z}{3})^3$ The initial ineq is equivalent to $\prod (\frac{x+y}{2})^4 \ge x^3y^3z^3(\frac{x+y+z}{3})^3$ if we let $x=s-a$ and so on. So popoviciu + amgm = solution

Soarer wrote: perfect_radio wrote: 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 ? Try popoviciu on f(x) = ln x You can get $\prod (\frac{x+y}{2})^2 \ge xyz(\frac{x+y+z}{3})^3$ The initial ineq is equivalent to $\prod (\frac{x+y}{2})^4 \ge x^3y^3z^3(\frac{x+y+z}{3})^3$ if we let $x=s-a$ and so on. So popoviciu + amgm = solution or we can just use the fact that $S= \frac{abc}{4R}$ and use $R \geq \frac{2p \sqrt 3}{9}$ where $p= \frac{a+b+c}{2}$ and $R$ is the radius of the circumscribed circle.

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

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 $A$!]

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 $$\tan{\frac{A}{2}}+\tan{\frac{B}{2}}+\tan{\frac{C}{2}} \ge \sqrt{4-8\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}}.$$

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

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.

Interesting

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

$\blacksquare$

There's ma solution

$$a^{2} + b^{2} + c^{2}\geq 4S \sqrt {3}$$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

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

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. $\square$.

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.