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Question Number 3832 by Yozzii last updated on 21/Dec/15

S h o w t h a t, \forall a, b \in R^{+},

{(\frac{a}{b})}^{1 / 3} + {(\frac{b}{a})}^{1 / 3} ⩽ {2 (a + b) (\frac{1}{a} + \frac{1}{b})}^{1 / 3} .

Commented by RasheedSindhi last updated on 22/Dec/15

{(\frac{a}{b})}^{1 / 3} + {(\frac{b}{a})}^{1 / 3} ⩽ {2 (a + b) (\frac{1}{a} + \frac{1}{b})}^{1 / 3}

R H S : {2 (a + b) (\frac{1}{a} + \frac{1}{b})}^{1 / 3}

= {2 (a + b) (\frac{a + b}{a b})}^{1 / 3}

= {\frac{2 {(a + b)}^{2}}{a b}}^{1 / 3} = \frac{2^{1 / 3} {(a + b)}^{2 / 3}}{{(a b)}^{1 / 3}}

L H S : {(\frac{a}{b})}^{1 / 3} + {(\frac{b}{a})}^{1 / 3}

= \frac{a^{1 / 3}}{b^{1 / 3}} + \frac{b^{1 / 3}}{a^{1 / 3}} = \frac{a^{2 / 3} + b^{2 / 3}}{{(a b)}^{1 / 3}}

L H S ⩽ R H S

\frac{a^{2 / 3} + b^{2 / 3}}{{(a b)}^{1 / 3}} ⩽ \frac{2^{1 / 3} {(a + b)}^{2 / 3}}{{(a b)}^{1 / 3}}

\Rightarrow a^{2 / 3} + b^{2 / 3} ⩽ 2^{1 / 3} {(a + b)}^{2 / 3}

[∵ a, b \in R^{+}]

\Rightarrow a^{2 / 3} + b^{2 / 3} ⩽ {(\sqrt{2})}^{2 / 3} {(a + b)}^{2 / 3}

\Rightarrow a^{2 / 3} + b^{2 / 3} ⩽ {(\sqrt{2} a + \sqrt{2} b)}^{2 / 3}

C a s e (i) a = b

a^{2 / 3} + b^{2 / 3} ⩽ {(\sqrt{2} a + \sqrt{2} b)}^{2 / 3}

\Rightarrow 2 a^{2 / 3} ⩽ {(2 \sqrt{2} a)}^{2 / 3}

⋮

C a s e (i i) a > b

a^{2 / 3} + b^{2 / 3} ⩽ {(\sqrt{2} a + \sqrt{2} b)}^{2 / 3}

\Rightarrow a^{2 / 3} (1 + {(\frac{b}{a})}^{2 / 3}) ⩽ {(\sqrt{2} a)}^{2 / 3} {1 + \frac{b}{a}}^{2 / 3}

\Rightarrow a^{2 / 3} (1 + {(\frac{b}{a})}^{2 / 3}) ⩽ 2^{1 / 3} a^{2 / 3} {1 + \frac{b}{a}}^{2 / 3}

D i v i d i n g b y a^{2 / 3} (> 0)

\Rightarrow (1 + {(\frac{b}{a})}^{2 / 3}) ⩽ 2^{1 / 3} {1 + \frac{b}{a}}^{2 / 3}

U s i n g b i n o m i a l t h e o r m

{(1 + x)}^{n} = 1 + n x + \frac{n (n - 1)}{2!} x^{2} + \dots

w h e r e ∣ x ∣< 1

{(1 + \frac{b}{a})}^{2 / 3} = 1 + (\frac{2}{3}) {(\frac{b}{a})}^{2 / 3} + \dots

C o n t i n u e

Commented by Yozzii last updated on 22/Dec/15

{T h e r e}^{'} s a n e r r o r i n t h e t h i r d l i n e

o f y o u r a t t e m p t .

^{'} \dots = {\frac{2 {(a + b)}^{2}}{a b}}^{1 / 3} = \frac{2^{1 / 3} {(a + b)}^{2 / 3}}{{(a b)}^{1 / 3}} \dots^{'}

Commented by Rasheed Soomro last updated on 22/Dec/15

T h a n k S! I a m g o i n g t o c o r r e c t i t .

Answered by Rasheed Soomro last updated on 23/Dec/15

S h o w t h a t, \forall a, b \in R^{+},

{(\frac{a}{b})}^{1 / 3} + {(\frac{b}{a})}^{1 / 3} ⩽ {2 (a + b) (\frac{1}{a} + \frac{1}{b})}^{1 / 3} \dots \dots \dots \dots \dots \dots (I)

- - - - - - - - - - - - - - -

A ⩽ B \Leftrightarrow A^{3} ⩽ B^{3}, \forall A, B > 0

{(\frac{a}{b})}^{1 / 3} + {(\frac{b}{a})}^{1 / 3} ⩽ {2 (a + b) (\frac{1}{a} + \frac{1}{b})}^{1 / 3}

\Rightarrow {[{(\frac{a}{b})}^{1 / 3} + {(\frac{b}{a})}^{1 / 3}]}^{3} ⩽ {[{2 (a + b) (\frac{1}{a} + \frac{1}{b})}^{1 / 3}]}^{3}

\Rightarrow \frac{a}{b} + 3 {(\frac{a}{b})}^{1 / 3} {(\frac{b}{a})}^{1 / 3} {{(\frac{a}{b})}^{1 / 3} + {(\frac{b}{a})}^{1 / 3}} + \frac{b}{a}

⩽ 2 (a + b) (\frac{a + b}{a b})

\Rightarrow \frac{a}{b} + 3 ({(\frac{a}{b})}^{1 / 3} + {(\frac{b}{a})}^{1 / 3}) + \frac{b}{a} ⩽ \frac{2 {(a + b)}^{2}}{a b}

\frac{a^{2} + b^{2}}{a b} + 3 (\frac{a^{2 / 3} + b^{2 / 3}}{a^{1 / 3} b^{1 / 3}}) ⩽ 2 (\frac{a^{2} + b^{2}}{a b}) + \frac{4 a b}{a b}

3 (\frac{a^{2 / 3} + b^{2 / 3}}{a^{1 / 3} b^{1 / 3}}) ⩽ \frac{a^{2} + b^{2}}{a b} + 4

3 (\frac{a^{2 / 3} + b^{2 / 3}}{a^{1 / 3} b^{1 / 3}}) a b ⩽ a^{2} + b^{2} + 4 a b [M u l t i p l y i n g b y a b > 0]

3 (a^{2 / 3} + b^{2 / 3}) a^{2 / 3} b^{2 / 3} ⩽ a^{2} + b^{2} + 4 a b

C o n t i n u e

Commented by Yozzii last updated on 24/Dec/15

T h a n k s f o r t r y i n g t h i s q u e s t i o n .

I t w a s t a k e n f r o m a n a t i o n a l m a t h

o l y m p i a d e x a m (C z e c h a n d S l o v a k

R e p u b l i c) i n t h e y e a r 2000 .

I {h a v e n}^{'} t g o t t e n a n y w h e r e n e a r s t a r t i n g

t o s o l v e t h e p r o b l e m .

Commented by Rasheed Soomro last updated on 24/Dec/15

M y E n g l i s h i s p o o r . S o I

C {o u l d n}^{'} t u n d e r s t a n d s e n s e o f l a s t

t w o l i n e s . P l e x p l a i n .

Commented by Yozzii last updated on 24/Dec/15

T h e p r o b l e m i s s o d i f f i c u l t f o r m e

t h a t, i t i s t a k i n g a l o t o f t i m e f o r m e

t o s o l v e i t .

Commented by Rasheed Soomro last updated on 24/Dec/15

T han k S!

Answered by Rasheed Soomro last updated on 23/Dec/15

\Rightarrow 1 + {(\frac{b}{a})}^{2 / 3} ⩽ 2^{1 / 3} {(1 + \frac{b}{a})}^{2 / 3} [∵ a^{2 / 3} > 0, d i v i d i n g b y i t]

{(\frac{a}{b})}^{1 / 3} + {(\frac{b}{a})}^{1 / 3} ⩽ {2 (a + b) (\frac{1}{a} + \frac{1}{b})}^{1 / 3}

\frac{a^{1 / 3}}{b^{1 / 3}} + \frac{b^{1 / 3}}{a^{1 / 3}} ⩽ {2 (a + b) (\frac{a + b}{a b})}^{1 / 3}

\frac{a^{2 / 3} + b^{2 / 3}}{{(a b)}^{1 / 3}} ⩽ {\frac{2 {(a + b)}^{2}}{a b}}^{1 / 3}

\frac{a^{2 / 3} + b^{2 / 3}}{{(a b)}^{1 / 3}} ⩽ \frac{2^{1 / 3} {(a + b)}^{2 / 3}}{{(a b)}^{1 / 3}}

a^{2 / 3} + b^{2 / 3} ⩽ 2^{1 / 3} {(a + b)}^{2 / 3} (i) [∵ {(a b)}^{1 / 3} > 0, s o b y m u l t i p l y i n g]

C a s e (I) a = b

(i) w i l l b e

2 a^{2 / 3} ⩽ 2^{1 / 3} {(2 a)}^{2 / 3}

2 a^{2 / 3} ⩽ 2^{1 / 3} 2^{2 / 3} a^{2 / 3}

2 a^{2 / 3} ⩽ 2 a^{2 / 3}

∴ ⩽ h o l d s

C a s e (I I) a > b

(i) w i l l b e

a^{2 / 3} (1 + {(\frac{b}{a})}^{2 / 3}) ⩽ 2^{1 / 3} a^{2 / 3} {(1 + \frac{b}{a})}^{2 / 3}

\Rightarrow 1 + {(\frac{b}{a})}^{2 / 3} ⩽ 2^{1 / 3} {(1 + \frac{b}{a})}^{2 / 3} [∵ a^{2 / 3} > 0, d i v i d i n g b y i t]

\Rightarrow 1 + {(\frac{b^{2}}{a^{2}})}^{1 / 3} ⩽ 2^{1 / 3} {{(1 + \frac{b}{a})}^{2}}^{1 / 3}

\Rightarrow 1 + {(\frac{b^{2}}{a^{2}})}^{1 / 3} ⩽ 2^{1 / 3} {(1 + \frac{2 b}{a} + \frac{b^{2}}{a^{2}})}^{1 / 3}

L e t \frac{b}{a} = x t h e n f o r x < 1

1 + x^{2 / 3} ⩽ 2^{1 / 3} {(1 + 2 x + x^{2})}^{1 / 3}

O r 1 + x^{2 / 3} ⩽ 2^{1 / 3} {(1 + x)}^{2 / 3}

I t {c o u l d n}^{'} t b e s o l v e d b y m e .

Commented by RasheedSindhi last updated on 25/Dec/15

I n 4^{t h} l i n e w h e r e d o e s \frac{2}{3} d i s a p p e a r ?

Commented by prakash jain last updated on 25/Dec/15

1 + x^{2 / 3} ⩽ 2^{1 / 3} {(1 + x)}^{2 / 3}

f (x) = 2^{1 / 3} {(1 + x)}^{2 / 3} - 1 - x^{2 / 3}

f^{'} (x) = 2^{1 / 3} \frac{2}{3} {(1 + x)}^{- 1 / 3} - \frac{2}{3} x^{- 1 / 3}

= \frac{2}{3} (2^{1 / 3} {(1 + x)}^{- 1 / 3} - x^{- 1 / 3})

\frac{2}{1 + x} - \frac{1}{x} = \frac{2 x - 1 - x}{x (1 + x)} < 0 for 0 < x < 1

\frac{2}{1 + x} < \frac{1}{x} \Rightarrow 2^{1 / 3} {(1 + x)}^{- 1 / 3} < x^{- 1 / 3}

f^{'} (x) ⩽ 0 0 < x < 1

f (x) is decreasing .

f (0) = 2^{1 / 3} - 1 ⩾ 0

f (1) = 0

Since f (x) is strictly decreasing for 0 < x < 1

and f (0) > 0 and f (1) = 0 f (x) > 0 f o r 0 < x < 1 .

b > a n e e d n o t b e c h e c k e d s i n c e i n e q u a l i t y

i s s y m m e t r i c i n a a n d b .

x = b = 0 LHS and RHS are undefined .

Commented by prakash jain last updated on 25/Dec/15

We are only interested in sign .

Commented by Rasheed Soomro last updated on 25/Dec/15

T h α n k S!