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Question Number 13236 by Abbas-Nahi last updated on 17/May/17

F o r p o s i t i v e a, b, c s u c h t h a t a b c = 1

s h o w t h a t a^{b + c} b^{c + a} c^{a + b} ⩽ 1

s o l u t i o n :

a^{b + c} b^{c + a} c^{a + b} = (a^{b} a^{c} b^{c} b^{a} c^{a} c^{b})

= {(b \times c)}^{a} {(a \times c)}^{b} {(a \times b)}^{c}

= {(a^{0} \times b \times c)}^{a} {(a \times b^{0} \times c)}^{b} {(a \times b \times c^{0})}^{c}

⩽ {(a \times b \times c)}^{a} {(a \times b^{} \times c)}^{b} {(a \times b \times c)}^{c}

⩽ {(1)}^{a} {(1)}^{b} {(1)}^{c}; s i n c e a b c = 1

⩽ 1

Commented by RasheedSindhi last updated on 17/May/17

= {(a^{0} \times b \times c)}^{a} {(a \times b^{0} \times c)}^{b} {(a \times b \times c^{0})}^{c}

⩽ {(a \times b \times c)}^{a} {(a \times b^{} \times c)}^{b} {(a \times b \times c)}^{c}

?

Do a^{a} b^{b} c^{c} ⩾ 1 ?

Commented by mrW1 last updated on 17/May/17

h o w d i d y o u g e t

{(a^{0} \times b \times c)}^{a}

⩽ {(a \times b \times c)}^{a}

?

s i n c e i f a < 1, w e h a v e a^{a} < 1 a n d

{(a^{0} \times b \times c)}^{a} > {(a \times b \times c)}^{a}

Commented by Tinkutara last updated on 17/May/17

Yes, your solution is incorrect . Please

check it . I now found a better solution .

From the given expression

a^{b + c} b^{c + a} c^{a + b} = a^{b + c} {(\frac{1}{a c})}^{c + a} c^{a + b} [∵ a b c = 1]

= a^{b + c} a^{- (c + a)} c^{- (c + a)} c^{a + b}

= \frac{a^{b + c - c - a}}{c^{c + a - a - b}} = \frac{a^{b - a}}{c^{c - b}} .

We can assume that a < b < c as the

equation is symmetric .

So the numbers are positive and raised

to positive powers .

So the denominator is greater than the

numerator .

Hence the above expression is ⩽ 1 .

Commented by RasheedSindhi last updated on 17/May/17

Mr . Tinkutara

“ So the numbers are positive {\underline{integers}}^{″}

The given codition is only

a, b, c > 0 \land a b c = 1

Why {you}^{'} ve assumed a, b, c as

integers ?

Commented by Abbas-Nahi last updated on 17/May/17

s o r r y s i r s i n c e a^{0} ⩽ a^{1} t h e n

{(a^{0} \times b \times c)}^{a} ⩽ {(a^{1} \times b \times c)}^{a}

(b y c o m p a r i n g)

Commented by Abbas-Nahi last updated on 17/May/17

t h e n {(a^{0} \times b \times c)}^{a} ⩽ (1)

s i n c e {(a^{1} \times b \times c)}^{a} = {(1)}^{a} = 1

Commented by Abbas-Nahi last updated on 17/May/17

T o M r . R a s h e e d

m y s i r s i n c e a^{0} ⩽ a^{1}, b^{0} ⩽ b^{1}, c^{0} ⩽ c^{1} t h e n

{(a^{0} \times b \times c)}^{a} {(a \times b^{0} \times c)}^{b} {(a \times b \times c^{0})}^{c}

⩽ {(a \times b \times c)}^{a} {(a \times b \times c)}^{b} {(a \times b \times c)}^{c}

(c o m p a r i n g) a n d s i n c e (a \times b \times c = 1)

⩽ {(1)}^{a} {(1)}^{b} {(1)}^{c}

⩽ (1) (1) (1)

⩽ (1)

f i n i s h e d

Commented by RasheedSindhi last updated on 19/May/17

But not necessarily

a^{0} ⩽ a^{1}

For example if a = 1 / 2

{(1 / 2)}^{0} \overset{?}{⩽} {(1 / 2)}^{1}

1 ⪇ \frac{1}{2}

Actually for any positive a

less than 1 :

a^{0} ⪇ a^{1}

Commented by Abbas-Nahi last updated on 17/May/17

s o r r y s i r; t h e q u e s t i o n a n d t h e r e l a t i o n

a b o v e a r e T r u e i f a n d o n l y i f a = 0 \land a = 1

Commented by mrW1 last updated on 17/May/17

i f a = 0 \Rightarrow a b c = 0, b u t a b c = 1!

Commented by Tinkutara last updated on 17/May/17

We cannot take a = 0 because it will

violate the condition a b c = 1 .

Commented by RasheedSindhi last updated on 17/May/17

a = 0 \land a = 1 means a is simultaneously

equal to 0 & 1 ?

Given condition : a b c = 1

a, b, c may take any real value

other than 0 .

For example :

(1 / 2) (6) (1 / 3)

To Mr . Abbas

pl set a = 1 / 2, b = 6 and c = 1 / 3

in your proof and see!

(These values fulfil a b c = 1)

Commented by Tinkutara last updated on 17/May/17

Sorry, it will not be integers .

Commented by prakash jain last updated on 18/May/17

To Tinkutara

Your argument that positive

numbers raised to + ve power

hence denominator is greater

than numerator is not valid .

In general

a > b, x > 0, y > 0

does not imply

a^{x} > b^{y}

Commented by Tinkutara last updated on 19/May/17

But here it is given that a b c = 1 .

Commented by prakash jain last updated on 19/May/17

I added a clarification

that the readers know that an

additional condition is required

a < c is not sufficient .

MrW 1 added in another post

that a ⩽ 1 and c ⩾ 1 which is an

additional condition .