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Consider the points $(a,0), (0,b), (0,c)$ in the $xy$-plane, where $a,b,c>0$ and $c>b$, and $r$ is the length of the line from $(0,0)$ to $(a,c)$, and $s$ is the length of the line from $(0,0)$ to $(a,b)$. How to show that $$\frac{c}{r}\geq \frac{b}{s}$$

I don't see any use of similar of triangles idea, so what is the key idea here?

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up vote 1 down vote accepted

Note that $r=\sqrt{a^2+c^2}$ and $s=\sqrt {a^2+b^2}$. You are asking to prove $\frac cr \ge \frac bs$, which is the same as $\frac c{\sqrt{a^2+c^2}} \ge \frac b{\sqrt {a^2+b^2}}$, which is the same as $\frac 1{\sqrt{1+\frac{a^2}{c^2}}} \ge \frac 1{\sqrt{1+\frac{a^2}{b^2}}}$. The smaller denominator in $\frac {a^2}{b^2}$ means the square root is larger, which means that its inverse is smaller. In fact, since $c \gt b$ the inequality is strict.

Alternately, you can use that $\frac cr$ is the cosine of the angle from the $+y$ axis to the segment from $(0,0)$ to $(a,c)$. Similarly $\frac bs$ is the cosine of the angle from the $+y$ axis to the line segment from $(0,0)$ to $(a,b)$. Since the first angle is smaller and cosine is a decreasing function, you have the inequality you want.

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That was very helpful, especially the second proof, thanks! – Monica L. Sep 28 '12 at 20:27

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