Let-PQRS-be-a-rectangle-such-that-PQ-a-and-QR-b-Suppose-r-1-is-the-radius-of-the-circle-passing-through-P-and-Q-and-touching-RS-and-r-2-is-the-radius-of-the-circle-passing-through-Q-and-R-and-t

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

Let PQRS be a rectangle such that

PQ = a and QR = b . Suppose r_{1} is the

radius of the circle passing through P

and Q and touching RS and r_{2} is the

radius of the circle passing through Q

and R and touching PS . Show that :

5 (a + b) ⩽ 8 (r_{1} + r_{2})

Commented by Tinkutara last updated on 02/Aug/17

Thank you very much ajfour Sir!

Commented by ajfour last updated on 02/Aug/17

Commented by ajfour last updated on 02/Aug/17

C_{1} \equiv [b - r_{1}, \frac{a}{2}] C_{2} \equiv [\frac{b}{2}, a - r_{2}]

{QC}_{1}^{2} = r_{1}^{2} = {(b - r_{1})}^{2} + \frac{a^{2}}{4} \dots . (i)

and {QC}_{2}^{2} = r_{2}^{2} = \frac{b^{2}}{4} + {(a - r_{2})}^{2} \dots (ii)

from (i) :

b ({2 r}_{1} - b) = \frac{a^{2}}{4}

\Rightarrow {2 r}_{1} = b + \frac{a^{2}}{4 b} or {8 r}_{1} = 4 b + \frac{a^{2}}{b}

and from (ii) :

a ({2 r}_{2} - a) = \frac{b^{2}}{4} or {8 r}_{2} = 4 a + \frac{b^{2}}{a}

adding these to get :

8 (r_{1} + r_{2}) = 4 (a + b) + (\frac{a^{2}}{b} + \frac{b^{2}}{a})

It remains to prove that

\frac{a^{2}}{b} + \frac{b^{2}}{a} ⩾ a + b

or \frac{a^{3} + b^{3}}{ab} ⩾ a + b

or a^{3} - a^{2} b + b^{3} - {ab}^{2} ⩾ 0

or a^{2} (a - b) - b^{2} (a - b) ⩾ 0

(a^{2} - b^{2}) (a - b) ⩾ 0

(a + b) {(a - b)}^{2} ⩾ 0

which is true since a > 0, b > 0 .

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