Difference between revisions of "1961 IMO Problems/Problem 2"

Revision as of 10:37, 28 December 2007

Let $a$ , $b$ , and $c$ be the lengths of a triangle whose area is S. Prove that

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

In what case does equality hold?

Substitute $S=\sqrt{s(s-a)(s-b)(s-c)}$ , where $s=\frac{a+b+c}{2}$

This shows that the inequality is equivalent to $a^2b^2+b^2c^2+c^2a^2\le a^4+b^4+c^4$ .

This can be proven because $a^2b^2\le\frac{a^4+b^4}{2}$ . The equality holds when $a=b=c$ , or when the triangle is equilateral.

1961 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
All IMO Problems and Solutions

@@ Line 11: / Line 11: @@
 Substitute <math>S=\sqrt{s(s-a)(s-b)(s-c)}</math>, where <math>s=\frac{a+b+c}{2}</math>
-This shows that the inequality is equivalent to <math>a^2b^2+b^2c^2+c^2a^2\lea^4+b^4+c^4</math>.
+This shows that the inequality is equivalent to <math>a^2b^2+b^2c^2+c^2a^2\le a^4+b^4+c^4</math>.
 This can be proven because <math>a^2b^2\le\frac{a^4+b^4}{2}</math>. The equality holds when <math>a=b=c</math>, or when the triangle is equilateral.