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As I remember I have already posted the solution.Try to find it.

Armo wrote: As I remember I have already posted the solution.Try to find it. Please repost it. It is probably lost due to the attack on the forum.

Armo wrote: As I remember I have already posted the solution.Try to find it. Your solution has been lost because of the hacker attack on MathLinks, so it would be very nice if you post it again. Meanwhile, here is my solution: The fact that $O_1O_2 = r$ is the radius of the circles $k_1$ and $k_2$ implies that the point $O_1$ lies on the circle $k_2$, and the point $O_2$ lies on the circle $k_1$. Since the points A and B are symmetric to each other with respect to the line $O_1O_2$, the midpoint M of the segment AB lies on this line $O_1O_2$, and the line AB is perpendicular to the line $O_1O_2$. Now, since the point M is the midpoint of the segment AB, the segment PM is a median in triangle APB, and thus, by the well-known formula for the length of a median in a triangle, we have $4\cdot PM^2 = 2\cdot(PA^2+PB^2)-AB^2)$. Thus, $2\cdot(PA^2+PB^2) = 4\cdot PM^2 + AB^2$. Now, we have to prove that $PA^2+PB^2 \geq 2r^2$, i. e. that $2\cdot(PA^2+PB^2) \geq 4r^2$. In light of the equation $2\cdot(PA^2+PB^2) = 4\cdot PM^2 + AB^2$, this becomes equivalent to $4\cdot PM^2 + AB^2 \geq 4r^2$. But proving $4\cdot PM^2 + AB^2 \geq 4r^2$ is really easy: Among all points on the circle $k_2$, the point $O_1$ is the one nearest to the point M, since actually the point $O_1$ is the point where the segment $MO_2$ meets $k_2$. Hence, since P is a point on the circle $k_2$, we have $PM\geq O_1M$. Now, since the line AB is perpendicular to the line $O_1O_2$, the triangle $AMO_1$ is right-angled; the Pythagoras theorem, applied to this triangle, yields $O_1M^2 + AM^2 = O_1A^2$. Since $O_1A$ is clearly the radius r of the circle $k_1$, we thus have $O_1M^2 + AM^2 = r^2$. Since $PM\geq O_1M$, we thus have $PM^2 + AM^2 \geq r^2$. Multiplying this inequality by 4, we obtain $4\cdot PM^2 + 4\cdot AM^2 \geq 4r^2$. Now, $4\cdot AM^2 = (2\cdot AM)^2 = AB^2$ (since $AB=2\cdot AM$ because of the point M being the midpoint of the segment AB); hence, we get $4\cdot PM^2 + AB^2 \geq 4r^2$. Proof complete. Darij

Nice proof darij. My proof is not nice but it is straightforward. As $ O_{1}O_{2}=r $ we have that $ O_{2} $ is on the circle $ k_{1} $. Take Cartesian coordinates system with origin $ O_{1} $ and abscissa axes $ O_{1}O_{2} $. Using our definations we get $ P(x;y), O_{2}(r;0), A(x_{0};y_{0}), B(x_{0};-y_{0}) $. So we have to prove that $ (x-x_{0})^{2}+(y-y_{0})^{2}+(x-x_{0})^{2}+(y+y_{0})^{2}\geq{2r^{2}}. $ But now as A,P lie on the circles $ k_{1},k_{2} $ respectively, we have $ x_{0}^{2}+y_{0}^{2}=r^{2}, (x-r)^{2}+y^{2}=r^{2} $ (from the circle equation). Putting this into our inequality we obtain $ 4xr\geq{4xx_{0}} $ which is obvious as $ x>0 $ and $ r\geq{x_{0}} $. QED