A Tricky Integral Inequality

First note that, due to Fubini's theorem, both repeated integrals can be treated as double. Next observe that the integrands are the (Euclidean) distance functions: $\sqrt{x^2+y^2-6x+9}=dist(B,P)\;$ and $\sqrt{x^2+y^2-8y+16}=dist(A,P),\;$ where $A,B,P\;$ are defined below:

By the triangle inequality then

$\displaystyle\begin{align} \Omega_1+\Omega_2&=\int_0^a\int_0^a\left(\sqrt{x^2+y^2-6x+9}+\sqrt{x^2+y^2-8y+16}\right)dxdy\\ &= \int_0^a\int_0^a(dist(B,P)+dist(A,P))dxdy\\ &\ge \int_0^a\int_0^adist(A,B)dxdy\\ &=\int_0^a\int_0^a5\\ &=5a^2. \end{align}$

The inequality just proved is always strict and can be improved for specific values of $a.\;$ For example, it is not hard to see that, for $a\le 0.5,\;$ $dist(B,P)+dist(A,P)\gt 6.\;$

The beauty of the problem is in the implied generality. Indeed, any distance function can be used in place of the Euclidean distance to make the problem even more intriguing. For example, the taxicab distance leads to the following inequality:

$\displaystyle\int_0^a\left(\int_0^a(|x-3|+|y|)dx\right)dy+\int_0^a\left(\int_0^a(|x|+|y-4|)dy\right)dx\ge 7a^2.$

Using the bounded distances disguises the problem even further. For example, define

$\displaystyle\begin{align}\Omega_1&=\int_0^a\left(\int_0^a\frac{\sqrt{x^2+y^2-6x+9}}{1+\sqrt{x^2+y^2-6x+9}}dx\right)dy\\ \Omega_2&=\int_0^a\left(\int_0^a\frac{\sqrt{x^2+y^2-8y+16}}{1+\sqrt{x^2+y^2-8y+16}}dy\right)dx. \end{align}$

Then $\displaystyle\Omega_1+\Omega_2\ge\frac{5}{6}a^2.$ For another example, if

$\displaystyle\begin{align}\Omega_1&=\int_0^a\left(\int_0^a\frac{|x-3|+|y|}{1+(|x-3|+|y|)}dx\right)dy\\ \Omega_2&=\int_0^a\left(\int_0^a\frac{|x|+|y-4|}{1+(|x|+|y-4|)}dy\right)dx, \end{align}$

then $\displaystyle\Omega_1+\Omega_2\ge\frac{7}{8}a^2.$

The problem from the Romanian Mathematical Magazine has been posted at the CutTheKnotMath facebook page by Dan Sitaru, with a solution by Ravi Prakash.