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Question Number 18967 by Tinkutara last updated on 02/Aug/17

$$\mathrm{Let}\mathrm{PQRS}\mathrm{be}\mathrm{a}\mathrm{rectangle}\mathrm{such}\mathrm{that}$$$$\mathrm{PQ}=a\mathrm{and}\mathrm{QR}=b.\mathrm{Suppose}{\mathrm{r}}_1\mathrm{is}\mathrm{the}$$$$\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{passing}\mathrm{through}\mathrm{P}$$$$\mathrm{and}\mathrm{Q}\mathrm{and}\mathrm{touching}\mathrm{RS}\mathrm{and}{\mathrm{r}}_2\mathrm{is}\mathrm{the}$$$$\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{passing}\mathrm{through}\mathrm{Q}$$$$\mathrm{and}\mathrm{R}\mathrm{and}\mathrm{touching}\mathrm{PS}.\mathrm{Show}\mathrm{that}:$$$$5(a+b)⩽8({\mathrm{r}}_1+{\mathrm{r}}_2)$$

Commented by Tinkutara last updated on 02/Aug/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{ajfour}\mathrm{Sir}!$$

Commented by ajfour last updated on 02/Aug/17

Commented by ajfour last updated on 02/Aug/17

$${\mathrm{C}}_1\equiv [\mathrm{b}-{\mathrm{r}}_1,\frac{\mathrm{a}}{2}]{\mathrm{C}}_2\equiv [\frac{\mathrm{b}}{2},\mathrm{a}-{\mathrm{r}}_2]$$$${\mathrm{QC}}_1^2={\mathrm{r}}_1^2={(\mathrm{b}-{\mathrm{r}}_1)}^2+\frac{{\mathrm{a}}^2}{4}....\left(\mathrm{i}\right)$$$$\mathrm{and}{\mathrm{QC}}_2^2={\mathrm{r}}_2^2=\frac{{\mathrm{b}}^2}{4}+{(\mathrm{a}-{\mathrm{r}}_2)}^2...\left(\mathrm{ii}\right)$$$$\mathrm{from}\left(\mathrm{i}\right):$$$$\mathrm{b}({2\mathrm{r}}_1-\mathrm{b})=\frac{{\mathrm{a}}^2}{4}$$$$\Rightarrow {2\mathrm{r}}_1=\mathrm{b}+\frac{{\mathrm{a}}^2}{4\mathrm{b}}\mathrm{or}{8\mathrm{r}}_1=4\mathrm{b}+\frac{{\mathrm{a}}^2}{\mathrm{b}}$$$$\mathrm{and}\mathrm{from}\left(\mathrm{ii}\right):$$$$\mathrm{a}({2\mathrm{r}}_2-\mathrm{a})=\frac{{\mathrm{b}}^2}{4}\mathrm{or}{8\mathrm{r}}_2=4\mathrm{a}+\frac{{\mathrm{b}}^2}{\mathrm{a}}$$$$\mathrm{adding}\mathrm{these}\mathrm{to}\mathrm{get}:$$$$8({\mathrm{r}}_1+{\mathrm{r}}_2)=4(\mathrm{a}+\mathrm{b})+(\frac{{\mathrm{a}}^2}{\mathrm{b}}+\frac{{\mathrm{b}}^2}{\mathrm{a}})$$$$\mathrm{It}\mathrm{remains}\mathrm{to}\mathrm{prove}\mathrm{that}$$$$\frac{{\mathrm{a}}^2}{\mathrm{b}}+\frac{{\mathrm{b}}^2}{\mathrm{a}}⩾\mathrm{a}+\mathrm{b}$$$$\mathrm{or}\frac{{\mathrm{a}}^3+{\mathrm{b}}^3}{\mathrm{ab}}⩾\mathrm{a}+\mathrm{b}$$$$\mathrm{or}{\mathrm{a}}^3-{\mathrm{a}}^2\mathrm{b}+{\mathrm{b}}^3-{\mathrm{ab}}^2⩾0$$$$\mathrm{or}{\mathrm{a}}^2(\mathrm{a}-\mathrm{b})-{\mathrm{b}}^2(\mathrm{a}-\mathrm{b})⩾0$$$$({\mathrm{a}}^2-{\mathrm{b}}^2)(\mathrm{a}-\mathrm{b})⩾0$$$$(\mathrm{a}+\mathrm{b}){(\mathrm{a}-\mathrm{b})}^2⩾0$$$$\mathrm{which}\mathrm{is}\mathrm{true}\mathrm{since}\mathrm{a}>0,\mathrm{b}>0.$$

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