Pages

Wednesday, May 12, 2010

Triangle Inequality

In a triangle $ABC$, let $a = BC, b = AC, c = AB$. For any point $M$ and real numbers $x,y,z$, show that

$(x+y+z)(xMA^2 + yMB^2 + zMC^2) \geq xyc^2 + xzb^2 + yza^2$

Solution
We shall show that the inequality above is equivalent to
$$(x \vec{MA} + y \vec{MB} + z \vec{MC})^2 \geq 0$$

Indeed, since:
$$2\vec{MA} \vec{MB} = MA^2 + MB^2 - c^2$$
so
$$2xy\vec{MA} \vec{MB} = xyMA^2 + xyMB^2 - xyc^2$$
$$2yz\vec{MB} \vec{MC} = yzMB^2 + yzMC^2 - yza^2$$
$$2zx\vec{MC} \vec{MA} = zxMC^2 + zxMA^2 - zxb^2$$

Adding them, we obtain:
$$RHS = \sum x(y+z)MA^2 - 2\sum xy \vec{MA} \vec{MB}$$

So
$$LHS - RHS = \sum x^2MA^2 + \sum x(y+z)MA^2 - RHS = \sum x^2 MA^2 + 2\sum xy \vec{MA} \vec{MB}$$
$$= (x \vec{MA} + y \vec{MB} + z \vec{MC})^2 \geq 0 $$

No comments:

Post a Comment