Question-15700

Question Number 15700 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 12/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 13/Jun/17

j, i s t h e i n c i r c l e o f A \overset{Δ}{B C} .

M N ∥ A C, K L ∥ A B, O P ∥ B C

O P = x, M N = y, K L = z, A B = c, B C = a, A C = b .

a > b > c > 1

\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots

s h o w t h a t :

1) \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 2

2) \frac{a}{x} + \frac{b}{y} + \frac{c}{z} ⩾ \frac{9}{2}

Answered by mrW1 last updated on 13/Jun/17

h_{a} = \frac{2 A}{a} with A = area of ABC

r = \frac{A}{s} with s = semiperimeter

\frac{r}{h_{a}} = \frac{a}{2 s} = \frac{a}{p} with p = perimeter

\frac{x}{a} = \frac{h_{a} - r}{h_{a}} = 1 - \frac{r}{h_{a}} = 1 - \frac{a}{p}

similarly

\frac{y}{b} = 1 - \frac{b}{p}

\frac{z}{c} = 1 - \frac{c}{p}

\Rightarrow \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 3 - \frac{1}{p} (a + b + c) = 3 - 1 = 2

\frac{a}{x} = \frac{1}{\frac{x}{a}} = \frac{1}{1 - \frac{a}{p}} > 1

\frac{b}{y} = \frac{1}{1 - \frac{b}{p}} > 1

\frac{c}{z} = \frac{1}{1 - \frac{c}{p}} > 1

\frac{a}{x} + \frac{b}{y} + \frac{c}{x} = \frac{1}{1 - \frac{a}{p}} + \frac{1}{1 - \frac{b}{p}} + \frac{1}{1 - \frac{c}{p}}

⩾ 3 \times \frac{1}{1 - \frac{1}{3}} = \frac{9}{2}

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 13/Jun/17

t h a n k s f o r y o u r c a r e . e r r o r f i x e d .

Commented by ajfour last updated on 13/Jun/17

\frac{a}{p} ⩾ \frac{1}{3} (S i r, s o m e p r o o f p l e a s e) .

Commented by mrW1 last updated on 13/Jun/17

I {didn}^{'} t mean that \frac{a}{p} ⩾ \frac{1}{3}, but I mean

minimum for \frac{1}{1 - \frac{a}{p}} + \frac{1}{1 - \frac{b}{p}} + \frac{1}{1 - \frac{c}{p}} is

\frac{9}{2} when \frac{a}{p} = \frac{b}{p} = \frac{c}{p} = \frac{1}{3} .

Here is the proof :

let x = \frac{a}{p}, y = \frac{b}{p}, z = \frac{c}{p}

x + y + z = 1

\frac{1}{1 - \frac{a}{p}} + \frac{1}{1 - \frac{b}{p}} + \frac{1}{1 - \frac{c}{p}} = \frac{1}{1 - x} + \frac{1}{1 - y} + \frac{1}{1 - z}

= \frac{1}{1 - x} + \frac{1}{1 - y} + \frac{1}{x + y} = f (x, y)

\frac{\partial f}{\partial x} = \frac{1}{{(1 - x)}^{2}} - \frac{1}{{(x + y)}^{2}}

\frac{\partial f}{\partial y} = \frac{1}{{(1 - y)}^{2}} - \frac{1}{{(x + y)}^{2}}

f (x, y) is minimum when \frac{\partial f}{\partial x} = 0 and \frac{\partial f}{\partial y} = 0

\frac{\partial f}{\partial x} = \frac{1}{{(1 - x)}^{2}} - \frac{1}{{(x + y)}^{2}} = 0

\Rightarrow 1 - x = x + y \dots (i)

\frac{\partial f}{\partial y} = \frac{1}{{(1 - y)}^{2}} - \frac{1}{{(x + y)}^{2}} = 0

\Rightarrow 1 - y = x + y \dots (ii)

from (i) and (ii)

x = y = \frac{1}{3}

z = 1 - (x + y) = \frac{1}{3}

f {(x, y)}_{\min .} = 3 \times \frac{1}{1 - \frac{1}{3}} = \frac{9}{2}

that it is a minimum not a maximum

can be checked with \frac{\partial^{2} f}{\partial x^{2}}, \frac{\partial^{2} f}{\partial y^{2}}, \frac{\partial^{2} f}{\partial x \partial y} .

Commented by ajfour last updated on 13/Jun/17

t h a n k y o u i m m e n s e l y S i r .

i^{'} l l n e e d t o s t u d y p a r t i a l d e r i v a t i v e s

a g a i n, f o r w h e n i d i d i w a s n o t

m a t u r e e n o u g h t h e n . .

Commented by mrW1 last updated on 13/Jun/17

once you said symmetry is beautiful .

symmetry is also a part of the nature .

if a function is symmetric over several

variables, its maximum or minimum

is always then when all variables

are equal . this gives me the direction,

the rest is how to prove it mathematically .

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