isi-entrance 2005 Q5

isi-entrance · India · solved Sine and Cosine Rules Prove an inequality or ordering relationship in a triangle

In a triangle with angles $P$, $Q$, $R$, let $\alpha$, $\beta$, $\gamma$ be the angles $\angle QCR = 2P$, $\angle QIR = Q + R$, $\angle QOR = P + Q/2 + R/2$ respectively. Show that $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} > \frac{1}{45}$.

Since $1/x$ is a convex function, by Jensen's inequality: $$\frac{1/2P + 1/(Q+R)}{2} > \frac{1}{(2P+Q+R)/2}$$ $$\Rightarrow 1/(2P) + 1/(Q+R) > \frac{4}{2P+Q+R}$$ Then: $$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} > \frac{4}{2P+Q+R} + \frac{2}{2P+Q+R} = \frac{6}{P+180} > \frac{6}{90+180} = \frac{6}{270} = \frac{1}{45}$$

In a triangle with angles $P$, $Q$, $R$, let $\alpha$, $\beta$, $\gamma$ be the angles $\angle QCR = 2P$, $\angle QIR = Q + R$, $\angle QOR = P + Q/2 + R/2$ respectively. Show that $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} > \frac{1}{45}$.

Paper Questions

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