Distance Inequality


Distance Inequality

Solution 1

$a,b,c\,$ that satisfy constraints lie in triangle with vertices $A=\left(\frac{3}{2},\frac{3}{2},0\right)\,$ $B=\left(\frac{3}{2},0,\frac{3}{2}\right),\,$ and $C=\left(0,\frac{3}{2},\frac{3}{2}\right).\,$ On the other hand, $a^2+b^2+c^2\,$ is the square of the distance to the origin. Thus the question reduces to finding the farthest from the origin point in $\Delta ABC.$

The nearest point to the origin is clearly the center of the triangle as it's the foot of the perpendicular from the origin to the plane $a+b+c=3.\,$ Imagine an expanding sphere with the center at the origin. After reaching the center of the triangle, it will intersect the plane in expanding circles centered at the center of $\Delta ABC.\,$ The last position the points on that circle satisfy the constraints is when the circles passes through the vertices of the triangle. At these points,


It follows that everywhere else in the triangle $a^2+b^2+c^2\le\frac{9}{2}.$

Just for the fun of it, note that without the constraint $a,b,c\in [0,\frac{3}{2}],\,$ the inquality to prove would be $a^2+b^2+c^2\le 9,\,$ with equality at points $(3,0,0),\,$ $(0,3,0),\,$ $(0,0,3).$

More interesting is, perhaps, the asymmetric case where $a,b,c\in [0,2].\,$ The inequality to prove appears to be $a^2+b^2+c^2\le 5,\,$ with equality at points $(2,1,0)\,$ and permutations.

Solution 2

There exist $\lambda_1,\lambda_2,\lambda_3\in [0,1]\,$ such that

$\displaystyle\begin{align} a &= \lambda_1\cdot 0+(1-\lambda_1)\frac{3}{2};\\ b &= \lambda_2\cdot 0+(1-\lambda_2)\frac{3}{2};\\ c &= \lambda_3\cdot 0+(1-\lambda_3)\frac{3}{2}. \end{align}$

The constraint rewrites as $\displaystyle\sum_{k=1}^3(1-\lambda_k)\frac{3}{2}=3,\;$ so that $\displaystyle\sum_{k=1}^3\lambda_k=1.\,$ Further, using Jensen's inequality,

$\displaystyle\begin{align} a^2+b^2+c^2 &= \sum_{k=1}^3\left[\lambda_k\cdot 0+(1-\lambda_k)\frac{3}{2}\right]^2\\ &\le 0^2\sum_{k=1}^3\lambda_k+\left(\frac{3}{2}\right)^2\sum_{k=1}^3(1-\lambda_k)\\ &=\frac{9}{4}\left(3-\sum_{k=1}^3\lambda_k\right)\\ &=\frac{9}{4}\cdot 2=\frac{9}{2}. \end{align}$


Distance inequality


The problem has been kindly posted at the CutTheKnotMath facebook page by Leo Giugiuc, with a comment "Almost new year happy. Beautiful and a little complicated." Solution 2 is by Marian Dinca. Illustration is by Nassim Nicholas Taleb.


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