How to prove this inequality by geometric methods?
$$2500 \pi-100<\sqrt{1 \cdot 199}+\sqrt{2 \cdot 198}+\cdots+\sqrt{99 \cdot 101}<2500\pi$$
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See https://math.stackexchange.com/questions/312263/help-with-the-inequality-sum-k-11006-sqrtk-cdot-2014-k5062-pi for a very similar problem. – Martin R Jan 05 '20 at 09:43
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Thanks. However, it is necessary to solve by geometric methods – Vertum Jan 05 '20 at 09:56
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Welcome to Math.SE! The community here prefers/expects questions to include something of what the asker knows about the problem. (What have you tried? Where did you get stuck? etc) This information helps answerers tailor their responses to best serve you, without wasting time (theirs or yours) telling you things you already know or talking over your head. (It also helps convince people that you aren't simply trying to get them to do your homework for you.) Since comments are easily overlooked, edit your question to add such details. – Blue Jan 05 '20 at 10:06
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2To start with I would draw a circle with diameter 200 and recall the power of a point theorem. Then I would try to play with Riemann sums... – user Jan 05 '20 at 10:24
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Thanks. It solves the problem – Vertum Jan 05 '20 at 10:37
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1See also: Prove the inequality $\sum_{k=1}^{2n-1}\sqrt{k(4n-k)}<\pi n^2$ for all natural $n$. – Martin Sleziak Jan 05 '20 at 11:41
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Note that $$\sqrt{k(200-k)}=\sqrt{100^2-(k-100)^2}\qquad(1\leq k\leq99)\ .$$ This means that you should look at the circular disc $(x-100)^2+y^2\leq100^2$. Your sum is a rectangle approximation to the area of the upper left quarter of this disc.
Christian Blatter
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Thank. I used the similar idea $\sqrt{k(200-k)}$ is the height lowered from a right angle in a right triangle. The hypotenuse relies on diameter $d=200$ – Vertum Jan 05 '20 at 10:44