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IMO Hk TST

Geometry Difficulty 7.3 National Olympiad, round 2 Prove it Hong Kong

Let aa, bb, cc be the sides of a triangle, and TT its area. Prove that
a2+b2+c2≥43T+(a−b)2+(b−c)2+(c−a)2. a^2 + b^2 + c^2 \ge 4\sqrt{3}T + (a-b)^2 + (b-c)^2 + (c-a)^2.

When does the equality hold?

Solution

Let ss be the semiperimeter of the triangle, and let x=s−ax = s - a, y=s−by = s - b and z=s−cz = s - c. Note that xx, yy, z>0z > 0 by the triangle inequality. Now,
a2+b2+c2≥43T+(a−b)2+(b−c)2+(c−a)2⇔2ab+2bc+2ca≥43s(s−a)(s−b)(s−c)+a2+b2+c2⇔2∑cyc(y+z)(x+z)≥43(x+y+z)xyz+(y+z)2+(z+x)2+(x+y)2⇔4(xy+yz+zx)≥43(x+y+z)xyz⇔(xy+yz+zx)2≥3((xy)(yz)+(yz)(zx)+(zx)(xy)). \begin{align*} & a^2 + b^2 + c^2 \ge 4\sqrt{3T} + (a-b)^2 + (b-c)^2 + (c-a)^2 \\ \Leftrightarrow & \quad 2ab + 2bc + 2ca \ge 4\sqrt{3s(s-a)(s-b)(s-c)} + a^2 + b^2 + c^2 \\ \Leftrightarrow & \quad 2 \sum_{\text{cyc}} (y+z)(x+z) \ge 4\sqrt{3(x+y+z)xyz} + (y+z)^2 + (z+x)^2 + (x+y)^2 \\ \Leftrightarrow & \quad 4(xy + yz + zx) \ge 4\sqrt{3(x+y+z)xyz} \\ \Leftrightarrow & \quad (xy + yz + zx)^2 \ge 3((xy)(yz) + (yz)(zx) + (zx)(xy)). \end{align*}

This is a well-known inequality (for example, the difference of the left-hand side and the right-hand side is 12∑cyc(xy−yz)2\frac{1}{2} \sum_{\text{cyc}} (xy - yz)^2). Equality holds when xy=yz=zxxy = yz = zx, i.e. a=b=ca = b = c.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.