Prove-that-if-p-gt-q-gt-0-and-x-0-then-1-p-x-p-p-1-1-1-q-x-q-q-1-1-

Question Number 1928 by Yozzi last updated on 24/Oct/15

Prove that, if p>q>0 and x≥0, then (1/p)((x^p /(p+1))−1)≥(1/q)((x^q /(q+1))−1).

P r o v e t h a t, i f p > q > 0 a n d x ⩾ 0, t h e n

\frac{1}{p} (\frac{x^{p}}{p + 1} - 1) ⩾ \frac{1}{q} (\frac{x^{q}}{q + 1} - 1) .

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

Prove that, if p>q>0 and x≥0, then (1/p)((x^p /(p+1))−1)≥(1/q)((x^q /(q+1))−1). −−−−−−−−−−−−−−−− A Try. Not much confident Let′s try to achieve simpler equivalent of the given statment,which is easy to prove. Let p=q+k, where k>0 (1/(q+k))((x^(q+k) /(q+k+1))−1)≥(1/q)((x^q /(q+1))−1) (1/(q+k))(((x^(q+k) −q−k−1)/(q+k+1)))≥(1/q)(((x^q −q−1)/(q+1))) ((x^(q+k) −q−k−1)/((q+k)(q+k+1)))≥((x^q −q−1)/(q(q+1))) ((x^(q+k) −q−k−1)/(q^2 +q+2qk+k^2 +k))≥((x^q −q−1)/(q^2 +q)) q>0 ∧ k>0 ⇒ 2qk+k^2 +k>0 Reduction of 2qk+k^2 +k (a +ve number) in the denominator of greater/equal side will leave no effect on the inequality. Hence, ((x^(q+k) −q−k−1)/(q^2 +q))≥((x^q −q−1)/(q^2 +q)) Multiplying by q^2 +q (>0) to both sides x^(q+k) −q−k−1≥x^q −q−1 x^(q+k) −k≥x^q Since k>0 x^(q+k) −k+k≥x^q x^q .x^k ≥x^q Given that x>0 or x=0 for x>0, x^q can be cancelled from both sides: x^k ≥1 Let′s prove for x≥1 first. x^k ≥1 is completely equivalent of the given statement and it is clearly true since x≥1 ∧ k>0 Hence given statement is true for x≥1 For x=0 the result can be proved directly. Now the problem is for x∈(0,1) .... Continue

P r o v e t h a t, i f p > q > 0 a n d x ⩾ 0, t h e n

\frac{1}{p} (\frac{x^{p}}{p + 1} - 1) ⩾ \frac{1}{q} (\frac{x^{q}}{q + 1} - 1) .

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

A T r y . N o t m u c h c o n f i d e n t

{L e t}^{'} s t r y t o a c h i e v e s i m p l e r e q u i v a l e n t o f

t h e g i v e n s t a t m e n t, w h i c h i s e a s y t o p r o v e .

L e t p = q + k, w h e r e k > 0

\frac{1}{q + k} (\frac{x^{q + k}}{q + k + 1} - 1) ⩾ \frac{1}{q} (\frac{x^{q}}{q + 1} - 1)

\frac{1}{q + k} (\frac{x^{q + k} - q - k - 1}{q + k + 1}) ⩾ \frac{1}{q} (\frac{x^{q} - q - 1}{q + 1})

\frac{x^{q + k} - q - k - 1}{(q + k) (q + k + 1)} ⩾ \frac{x^{q} - q - 1}{q (q + 1)}

\frac{x^{q + k} - q - k - 1}{q^{2} + q + 2 q k + k^{2} + k} ⩾ \frac{x^{q} - q - 1}{q^{2} + q}

q > 0 \land k > 0 \Rightarrow 2 q k + k^{2} + k > 0

R e d u c t i o n o f 2 q k + k^{2} + k (a + v e n u m b e r)

i n t h e d e n o m i n a t o r o f g r e a t e r / e q u a l s i d e

w i l l l e a v e n o e f f e c t o n t h e i n e q u a l i t y .

H e n c e,

\frac{x^{q + k} - q - k - 1}{q^{2} + q} ⩾ \frac{x^{q} - q - 1}{q^{2} + q}

M u l t i p l y i n g b y q^{2} + q (> 0) t o b o t h s i d e s

x^{q + k} - q - k - 1 ⩾ x^{q} - q - 1

x^{q + k} - k ⩾ x^{q}

S i n c e k > 0

x^{q + k} - k + k ⩾ x^{q}

x^{q} . x^{k} ⩾ x^{q}

G i v e n t h a t x > 0 o r x = 0

f o r x > 0, x^{q} c a n b e c a n c e l l e d f r o m b o t h s i d e s :

x^{k} ⩾ 1

{L e t}^{'} s p r o v e f o r x ⩾ 1 f i r s t .

x^{k} ⩾ 1 i s c o m p l e t e l y e q u i v a l e n t o f

t h e g i v e n s t a t e m e n t a n d i t i s c l e a r l y t r u e s i n c e

x ⩾ 1 \land k > 0

H e n c e g i v e n s t a t e m e n t i s t r u e f o r x ⩾ 1

F o r x = 0 t h e r e s u l t c a n b e p r o v e d d i r e c t l y .

N o w t h e p r o b l e m i s f o r x \in (0, 1)

\dots .

C o n t i n u e

Commented by prakash jain last updated on 24/Oct/15

Observation on Rasheed Comment ((x^(q+k) −q−k−1)/(q^2 +q+2qk+k^2 +k))≥((x^2 −q−1)/(q^2 +q)) is not equivalent to (if 2qk+k^2 +k>0) ((x^(q+k) −q−k−1)/(q^2 +q))≥((x^2 −q−1)/(q^2 +q)) since ((x^(q+k) −q−k−1)/(q^2 +q+2qk+k^2 +k)) < ((x^(q+k) −q−k−1)/(q^2 +q)) Suggestion: inequality may be simpler to prove if you try p=kq, k>1

Observation on Rasheed Comment

\frac{x^{q + k} - q - k - 1}{q^{2} + q + 2 q k + k^{2} + k} ⩾ \frac{x^{2} - q - 1}{q^{2} + q}

is not equivalent to (if 2 q k + k^{2} + k > 0)

\frac{x^{q + k} - q - k - 1}{q^{2} + q} ⩾ \frac{x^{2} - q - 1}{q^{2} + q}

since \frac{x^{q + k} - q - k - 1}{q^{2} + q + 2 q k + k^{2} + k} < \frac{x^{q + k} - q - k - 1}{q^{2} + q}

Suggestion : inequality may be simpler to prove

if you try p = k q, k > 1

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

THANKSSSSSsss... for correction and suggestion!

THANKS S S S S s s s \dots f o r c o r r e c t i o n a n d s u g g e s t i o n!

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

Prove that, if p>q>0 and x≥0, then (1/p)((x^p /(p+1))−1)≥(1/q)((x^q /(q+1))−1). −−−−−×××−−−−−−− To get a simpler equivalent(?): (x^p /(p(p+1)))−(1/p)≥(x^q /(q(q+1)))−(1/q)......................(1) Now, p>q ⇒ (1/p)<(1/q) ⇒−(1/p)>−(1/q)..........(2) Adding (1) and (2) (x^p /(p(p+1))) ≥ (x^q /(q(q+1))) Is this equivalent to the original? Does adding a same−sense inequality yield an equivalent inequality? Let p=qk, where k>1 (x^(qk) /(qk(qk+1))) ≥ (x^q /(q(q+1))) (x^(qk) /(k(qk+1))) ≥ (x^q /(q+1)) [Multiplying by q(>0) ] (x^(qk) /(qk^2 +k)) ≥ (x^q /(q+1)) Continue

P r o v e t h a t, i f p > q > 0 a n d x ⩾ 0, t h e n

\frac{1}{p} (\frac{x^{p}}{p + 1} - 1) ⩾ \frac{1}{q} (\frac{x^{q}}{q + 1} - 1) .

- - - - - \times \times \times - - - - - - -

T o g e t a simpler equivalent (?) :

\frac{x^{p}}{p (p + 1)} - \frac{1}{p} ⩾ \frac{x^{q}}{q (q + 1)} - \frac{1}{q} \dots \dots \dots \dots \dots \dots \dots . (1)

N o w,

p > q \Rightarrow \frac{1}{p} < \frac{1}{q} \Rightarrow - \frac{1}{p} > - \frac{1}{q} \dots \dots \dots . (2)

A d d i n g (1) a n d (2)

\frac{x^{p}}{p (p + 1)} ⩾ \frac{x^{q}}{q (q + 1)}

I s t h i s e q u i v a l e n t t o t h e o r i g i n a l ?

D o e s a d d i n g a same - sense inequality y i e l d a n

equivalent inequality ?

L e t p = q k, w h e r e k > 1

\frac{x^{q k}}{q k (q k + 1)} ⩾ \frac{x^{q}}{q (q + 1)}

\frac{x^{q k}}{k (q k + 1)} ⩾ \frac{x^{q}}{q + 1} [M u l t i p l y i n g b y q (> 0)]

\frac{x^{q k}}{{q k}^{2} + k} ⩾ \frac{x^{q}}{q + 1}

Continue

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

On the comment of prakash jain Sir you have said that a>b is not equivalent to A>B because (after dropping some +ve value from denominator) A<a. Does this mean a>b and A>B will be equivalent only when A=a?

On the comment of prakash jain

Sir y o u h a v e s a i d t h a t a > b i s n o t e q u i v a l e n t t o A > B

b e c a u s e (a f t e r d r o p p i n g s o m e + v e v a l u e f r o m d e n o m i n a t o r)

A < a .

D o e s t h i s m e a n a > b a n d A > B w i l l b e e q u i v a l e n t o n l y

w h e n A = a ?

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

While proving if a<A then a>b⇒ A>b but A>b⇏a>b if a>A then A>b⇒a>b but a>b⇏A>b

While proving

if a < A then a > b \Rightarrow A > b but A > b ⇏ a > b

if a > A then A > b \Rightarrow a > b but a > b ⇏ A > b

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

Rasheed regarding equality for p>q If (x^p /(p(p+1)))≥(x^q /(q(q+1))) ......(1) then (x^p /(p(p+1)))−(1/p)≥(x^q /(q(q+1))) −(1/q) ......(2) (1) ⇒(2) however (2) does not imply (1). Now you cannot prove 1 since is not true. Try x=1,p=2, q=1

Rasheed regarding equality for p > q

If \frac{x^{p}}{p (p + 1)} ⩾ \frac{x^{q}}{q (q + 1)} \dots \dots (1)

then \frac{x^{p}}{p (p + 1)} - \frac{1}{p} ⩾ \frac{x^{q}}{q (q + 1)} - \frac{1}{q} \dots \dots (2)

(1) \Rightarrow (2) however (2) does not imply (1) .

Now you cannot prove 1 since is not true .

Try x = 1, p = 2, q = 1

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

THANKS for so many valueable explanations! However I would like to understand clearly why (x^p /(p(p+1)))−(1/p)≥(x^q /(q(q+1)))−(1/q) does not imply_(−) to (x^p /(p(p+1)))≥(x^q /(q(q+1))) when p>q>0 is also given ? p>q ⇒(1/p)<(1/q)⇒−(1/p)>−(1/q) (x^p /(p(p+1)))−(1/p)≥(x^q /(q(q+1)))−(1/q) ∧ −(1/p)>−(1/q) ⇒(x^p /(p(p+1)))≥(x^q /(q(q+1))) [Subtracting second inequality from first] However as for as equivalence_(−) is concerned I admit that these two are not equivalent.Perhaps this is the reason that one satisfy some values and other not. But since the goal was to determine an equivalent so I failed to achieve. I think implication and equivalence are two different things. Anyway your opinion will be an expert′s opinion and my question is a student′s queztion. I am here to learn from you and other experts.

THANKS f o r s o m a n y v a l u e a b l e e x p l a n a t i o n s!

However I would like to understand clearly

why

\frac{x^{p}}{p (p + 1)} - \frac{1}{p} ⩾ \frac{x^{q}}{q (q + 1)} - \frac{1}{q}

does not \underline{imply} to

\frac{x^{p}}{p (p + 1)} ⩾ \frac{x^{q}}{q (q + 1)}

when p > q > 0 is also given ?

p > q \Rightarrow \frac{1}{p} < \frac{1}{q} \Rightarrow - \frac{1}{p} > - \frac{1}{q}

\frac{x^{p}}{p (p + 1)} - \frac{1}{p} ⩾ \frac{x^{q}}{q (q + 1)} - \frac{1}{q} \land - \frac{1}{p} > - \frac{1}{q}

\Rightarrow \frac{x^{p}}{p (p + 1)} ⩾ \frac{x^{q}}{q (q + 1)} [S u b t r a c t i n g s e c o n d i n e q u a l i t y f r o m f i r s t]

H o w e v e r a s f o r a s \underline{e q u i v a l e n c e} i s c o n c e r n e d I a d m i t

t h a t t h e s e t w o a r e n o t e q u i v a l e n t . P e r h a p s t h i s i s

t h e r e a s o n t h a t o n e s a t i s f y s o m e v a l u e s a n d o t h e r n o t .

B u t s i n c e t h e g o a l w a s t o d e t e r m i n e a n e q u i v a l e n t s o

I f a i l e d t o a c h i e v e .

I t h i n k i m p l i c a t i o n a n d e q u i v a l e n c e a r e t w o

d i f f e r e n t t h i n g s .

A n y w a y y o u r o p i n i o n w i l l b e a n {e x p e r t}^{'} s o p i n i o n

a n d m y q u e s t i o n i s a {s t u d e n t}^{'} s q u e z t i o n .

I a m h e r e t o l e a r n f r o m y o u a n d o t h e r e x p e r t s .

Commented by Rasheed Soomro last updated on 27/Oct/15

Prove that, if p>q>0 and x≥0, then (1/p)((x^p /(p+1))−1)≥(1/q)((x^q /(q+1))−1) −×−×−×−×−×−×−×−×−×− (x^p /(p(p+1)))−(1/p)≥(x^q /(q(q+1)))−(1/q) (x^p /(p(p+1)))−(x^q /(q(q+1)))≥(1/p)−(1/q) ((x^p q(q+1)−x^q p(p+1))/(pq(p+1)(q+1)))≥((q−p)/(pq)).....................(A) Approach(1) Let p=qk,where k>1_(−) ((x^p q(q+1)−x^q p(p+1))/(pq(p+1)(q+1)))≥((q−p)/(pq)).....................(A) ((x^(qk) q(q+1)−x^q qk(qk+1))/(q^2 k(qk+1)(q+1)))≥((q−qk)/(q^2 k)) ((x^(qk) q(q+1)−x^q qk(qk+1))/((qk+1)(q+1)))≥q(1−k) ((x^(qk) (q+1)−x^q k(qk+1))/((qk+1)(q+1)))≥1−k ((x^(qk) (q+1)−x^q k(qk+1))/((qk+1)(q+1)))+k≥1 ((x^(qk) (q+1)−x^q k(qk+1)+k(qk+1)(q+1))/((qk+1)(q+1))) ≥ 1 ((x^(qk) q+x^(qk) −x^q qk^2 −x^q k+q^2 k^2 +qk^2 +qk+k)/((qk+1)(q+1)))≥1 ((x^(qk) q+x^(qk) −x^q qk^2 −x^q k+q^2 k^2 +qk^2 +qk+k)/(q^2 k+qk+q+1))≥1 x^(qk) q+x^(qk) −x^q qk^2 −x^q k+q^2 k^2 +qk^2 +qk+k−q^2 k−qk−q−1≥0 ...... Approach(2) Let p=q+k,where k>0 ((x^p q(q+1)−x^q p(p+1))/(pq(p+1)(q+1)))≥((q−p)/(pq)).....................(A) ((x^((q+k)) q(q+1)−x^q (q+k){(q+k)+1})/(q(q+1){(q+k)+1}(q+1)))≥((q−(q+k))/(q(q+k))) ((x^(q+k) q(q+1)−x^q (q+k)(q+k+1))/(q(q+1)^2 (q+k+1)))≥((−k)/(q(q+k))) ((x^(q+k) q(q+1)−x^q (q+k)(q+k+1))/((q+1)^2 (q+k+1)))≥((−k)/(q+k)) [Multiply by q(>0)] ((x^(q+k) q(q+1)−x^q (q+k)(q+k+1))/((q+1)^2 (q+k+1)))+(k/(q+k)) ≥0 (({x^(q+k) q(q+1)−x^q (q+k)(q+k+1)}(q+k)+k(q+1)^2 (q+k+1))/((q+1)^2 (q+k)(q+k+1))) ≥0 Continue

P r o v e t h a t, i f p > q > 0 a n d x ⩾ 0, t h e n

\frac{1}{p} (\frac{x^{p}}{p + 1} - 1) ⩾ \frac{1}{q} (\frac{x^{q}}{q + 1} - 1)

- \times - \times - \times - \times - \times - \times - \times - \times - \times -

\frac{x^{p}}{p (p + 1)} - \frac{1}{p} ⩾ \frac{x^{q}}{q (q + 1)} - \frac{1}{q}

\frac{x^{p}}{p (p + 1)} - \frac{x^{q}}{q (q + 1)} ⩾ \frac{1}{p} - \frac{1}{q}

\frac{x^{p} q (q + 1) - x^{q} p (p + 1)}{p q (p + 1) (q + 1)} ⩾ \frac{q - p}{p q} \dots \dots \dots \dots \dots \dots \dots (A)

A p p r o a c h (1)

\underline{L e t p = q k, w h e r e k > 1}

\frac{x^{p} q (q + 1) - x^{q} p (p + 1)}{p q (p + 1) (q + 1)} ⩾ \frac{q - p}{p q} \dots \dots \dots \dots \dots \dots \dots (A)

\frac{x^{q k} q (q + 1) - x^{q} q k (q k + 1)}{q^{2} k (q k + 1) (q + 1)} ⩾ \frac{q - q k}{q^{2} k}

\frac{x^{q k} q (q + 1) - x^{q} q k (q k + 1)}{(q k + 1) (q + 1)} ⩾ q (1 - k)

\frac{x^{q k} (q + 1) - x^{q} k (q k + 1)}{(q k + 1) (q + 1)} ⩾ 1 - k

\frac{x^{q k} (q + 1) - x^{q} k (q k + 1)}{(q k + 1) (q + 1)} + k ⩾ 1

\frac{x^{q k} (q + 1) - x^{q} k (q k + 1) + k (q k + 1) (q + 1)}{(q k + 1) (q + 1)} ⩾ 1

\frac{x^{q k} q + x^{q k} - x^{q} {q k}^{2} - x^{q} k + q^{2} k^{2} + {q k}^{2} + q k + k}{(q k + 1) (q + 1)} ⩾ 1

\frac{x^{q k} q + x^{q k} - x^{q} {q k}^{2} - x^{q} k + q^{2} k^{2} + {q k}^{2} + q k + k}{q^{2} k + q k + q + 1} ⩾ 1

x^{q k} q + x^{q k} - x^{q} {q k}^{2} - x^{q} k + q^{2} k^{2} + {q k}^{2} + q k + k - q^{2} k - q k - q - 1 ⩾ 0

\dots \dots

A p p r o a c h (2)

L e t p = q + k, w h e r e k > 0

\frac{x^{p} q (q + 1) - x^{q} p (p + 1)}{p q (p + 1) (q + 1)} ⩾ \frac{q - p}{p q} \dots \dots \dots \dots \dots \dots \dots (A)

\frac{x^{(q + k)} q (q + 1) - x^{q} (q + k) {(q + k) + 1}}{q (q + 1) {(q + k) + 1} (q + 1)} ⩾ \frac{q - (q + k)}{q (q + k)}

\frac{x^{q + k} q (q + 1) - x^{q} (q + k) (q + k + 1)}{q {(q + 1)}^{2} (q + k + 1)} ⩾ \frac{- k}{q (q + k)}

\frac{x^{q + k} q (q + 1) - x^{q} (q + k) (q + k + 1)}{{(q + 1)}^{2} (q + k + 1)} ⩾ \frac{- k}{q + k} [M u l t i p l y b y q (> 0)]

\frac{x^{q + k} q (q + 1) - x^{q} (q + k) (q + k + 1)}{{(q + 1)}^{2} (q + k + 1)} + \frac{k}{q + k} ⩾ 0

\frac{{x^{q + k} q (q + 1) - x^{q} (q + k) (q + k + 1)} (q + k) + k {(q + 1)}^{2} (q + k + 1)}{{(q + 1)}^{2} (q + k) (q + k + 1)} ⩾ 0

C o n t i n u e

Commented by Rasheed Soomro last updated on 26/Oct/15

(1/p)((x^p /(p+1))−1)≥(1/q)((x^q /(q+1))−1) p>0,q>0⇒pq>0,multiplying by pq to both sides q((x^p /(p+1))−1)≥p((x^q /(q+1))−1) ((qx^p )/(p+1))−q≥((px^q )/(q+1))−p_ ((qx^p )/(p+1))+p≥((px^q )/(q+1))+q [Adding p+q to both sides] ((qx^p +p(p+1))/(p+1))≥((px^q +q(q+1))/(q+1)) Let p=qk, where k>1 ((qx^(qk) +qk(qk+1))/(qk+1))≥((qkx^q +q(q+1))/(q+1)) q>0 ∧ k>0⇒q+1>0 ∧ qk+1>0 ⇒(qk+1)(q+1)>0 Multiplying by (qk+1)(q+1) to both sides: q{x^(qk) +qk^2 +k}(q+1)≥q{kx^q +q+1}(qk+1) }÷q (>0) {x^(qk) +qk^2 +k}(q+1)≥{kx^q +q+1}(qk+1) qx^(qk) +q^2 k^2 +qk^(×) +x^(qk) +qk^2 +k≥qk^2 x^q +q^2 k+qk^(×) +kx^q +q+1 qx^(qk) +q^2 k^2 +x^(qk) +qk^2 +k≥qk^2 x^q +q^2 k+kx^q +q+1 (q+1)x^(qk) +q^2 k^2 +qk^2 +k≥k(qk+1)x^q +q^2 k+q+1 The above inequality is equivalent to the inequality to be proved.To prove the given it is suficient to prove its equivalent. If we could prove the following inequalities They imply aove inequality. Howdver vice versa is not correct. (q+1)x^(qk) ≥k(qk+1)x^q [.....].....................(i) q^2 k^2 ≥q^2 k⇒(q^2 k)k≥q^2 k [∵ k>1]..........(ii) qk^2 ≥q [k>1 ∧ q>0⇒k^2 >1⇒qk^2 ≥q.....(iii) k≥1 [Assumption]...........................(iv) Unfortinuately (i) is not always correct.For example at x=1 it is false! Dilli hunooz door ast Continue

\frac{1}{p} (\frac{x^{p}}{p + 1} - 1) ⩾ \frac{1}{q} (\frac{x^{q}}{q + 1} - 1)

p > 0, q > 0 \Rightarrow p q > 0, m u l t i p l y i n g b y p q t o b o t h s i d e s

q (\frac{x^{p}}{p + 1} - 1) ⩾ p (\frac{x^{q}}{q + 1} - 1)

\frac{{q x}^{p}}{p + 1} - q ⩾ \frac{{p x}^{q}}{q + 1} - p_{}

\frac{{q x}^{p}}{p + 1} + p ⩾ \frac{{p x}^{q}}{q + 1} + q [A d d i n g p + q t o b o t h s i d e s]

\frac{{q x}^{p} + p (p + 1)}{p + 1} ⩾ \frac{{p x}^{q} + q (q + 1)}{q + 1}

L e t p = q k, w h e r e k > 1

\frac{{q x}^{q k} + q k (q k + 1)}{q k + 1} ⩾ \frac{{q k x}^{q} + q (q + 1)}{q + 1}

q > 0 \land k > 0 \Rightarrow q + 1 > 0 \land q k + 1 > 0 \Rightarrow (q k + 1) (q + 1) > 0

M u l t i p l y i n g b y (q k + 1) (q + 1) t o b o t h s i d e s :

q {x^{q k} + {q k}^{2} + k} (q + 1) ⩾ q {{k x}^{q} + q + 1} (q k + 1)} \div q (> 0)

{x^{q k} + {q k}^{2} + k} (q + 1) ⩾ {{k x}^{q} + q + 1} (q k + 1)

{q x}^{q k} + q^{2} k^{2} + \overset{\times}{q k} + x^{q k} + {q k}^{2} + k ⩾ {q k}^{2} x^{q} + q^{2} k + \overset{\times}{q k} + {k x}^{q} + q + 1

{q x}^{q k} + q^{2} k^{2} + x^{q k} + {q k}^{2} + k ⩾ {q k}^{2} x^{q} + q^{2} k + {k x}^{q} + q + 1

(q + 1) x^{q k} + q^{2} k^{2} + {q k}^{2} + k ⩾ k (q k + 1) x^{q} + q^{2} k + q + 1

T h e a b o v e i n e q u a l i t y i s e q u i v a l e n t t o t h e i n e q u a l i t y

t o b e p r o v e d . T o p r o v e t h e g i v e n i t i s s u f i c i e n t t o

p r o v e i t s e q u i v a l e n t .

I f w e c o u l d p r o v e t h e f o l l o w i n g i n e q u a l i t i e s

T h e y i m p l y a o v e i n e q u a l i t y . H o w d v e r v i c e v e r s a

i s n o t c o r r e c t .

(q + 1) x^{q k} ⩾ k (q k + 1) x^{q} [\dots . .] \dots \dots \dots \dots \dots \dots \dots (i)

q^{2} k^{2} ⩾ q^{2} k \Rightarrow (q^{2} k) k ⩾ q^{2} k [∵ k > 1] \dots \dots \dots . (i i)

{q k}^{2} ⩾ q [k > 1 \land q > 0 \Rightarrow k^{2} > 1 \Rightarrow {q k}^{2} ⩾ q \dots . . (i i i)

k ⩾ 1 [A s s u m p t i o n] \dots \dots \dots \dots \dots \dots \dots \dots \dots (i v)

U n f o r t i n u a t e l y (i) i s n o t a l w a y s c o r r e c t . F o r e x a m p l e

a t x = 1 i t i s f a l s e!

D i l l i h u n o o z d o o r a s t

C o n t i n u e

Answered by Rasheed Soomro last updated on 26/Oct/15

Experiment (1/p)((x^p /(p+1))−1)≥(1/q)((x^q /(q+1))−1) Let p=q^k ,where k>1 (1/q^k )((x^q^k /(q^k +1))−1)≥(1/q)((x^q /(q+1))−1) (1/q^(k−1) )((x^q^k /(q^k +1))−1)≥((x^q /(q+1))−1) (x^q^k /(q^(q−1) (q^k +1)))−(1/q^(k−1) )≥((x^q −q−1)/(q+1)) ((x^q^k −q^k −1)/(q^(k−1) (q^k +1)))≥((x^q −q−1)/(q+1)) (x^q^k −q^k −1)(q+1)≥q^(k−1) (q^k +1)(x^q −q−1) qx^q^k −q^(k+1) −q+x^q^k −q^k −1≥q^(k−1) (q^k x^q −q^(k+1) −q^k +x^q −q−1) qx^q^k −q^(k+1) −q+x^q^k −q^k −1≥q^(2k−1) x^q −q^(2k) −q^(2k−1) +q^(k−1) x^q −q^k −q^(k−1) ) (q+1)x^q^k −q^(k+1) −q−q^k −1≥(q^(2k−1) +q^(k−1) )x^q −q^(2k) −q^(2k−1) −q^k −q^(k−1) )

E x p e r i m e n t

\frac{1}{p} (\frac{x^{p}}{p + 1} - 1) ⩾ \frac{1}{q} (\frac{x^{q}}{q + 1} - 1)

L e t p = q^{k}, w h e r e k > 1

\frac{1}{q^{k}} (\frac{x^{q^{k}}}{q^{k} + 1} - 1) ⩾ \frac{1}{q} (\frac{x^{q}}{q + 1} - 1)

\frac{1}{q^{k - 1}} (\frac{x^{q^{k}}}{q^{k} + 1} - 1) ⩾ (\frac{x^{q}}{q + 1} - 1)

\frac{x^{q^{k}}}{q^{q - 1} (q^{k} + 1)} - \frac{1}{q^{k - 1}} ⩾ \frac{x^{q} - q - 1}{q + 1}

\frac{x^{q^{k}} - q^{k} - 1}{q^{k - 1} (q^{k} + 1)} ⩾ \frac{x^{q} - q - 1}{q + 1}

(x^{q^{k}} - q^{k} - 1) (q + 1) ⩾ q^{k - 1} (q^{k} + 1) (x^{q} - q - 1)

{q x}^{q^{k}} - q^{k + 1} - q + x^{q^{k}} - q^{k} - 1 ⩾ q^{k - 1} (q^{k} x^{q} - q^{k + 1} - q^{k} + x^{q} - q - 1)

{q x}^{q^{k}} - q^{k + 1} - q + x^{q^{k}} - q^{k} - 1 ⩾ q^{2 k - 1} x^{q} - q^{2 k} - q^{2 k - 1} + q^{k - 1} x^{q} - q^{k} - q^{k - 1})

(q + 1) x^{q^{k}} - q^{k + 1} - q - q^{k} - 1 ⩾ (q^{2 k - 1} + q^{k - 1}) x^{q} - q^{2 k} - q^{2 k - 1} - q^{k} - q^{k - 1})

Answered by Rasheed Soomro last updated on 28/Oct/15

(1/p)((x^p /(p+1))−1)≥(1/q)((x^q /(q+1))−1) (x^p /(p(p+1)))−(1/p)≥(x^q /(q(q+1)))−(1/q) (x^p /(p(p+1)))−(x^q /(q(q+1)))≥(1/p)−(1/q) Let p=qk,where k>1 (x^(qk) /(qk(qk+1)))−(x^q /(q(q+1)))≥(1/(qk))−(1/q) (x^(qk) /(k(qk+1)))−(x^q /((q+1)))≥(1/k)−1 [Multkplying by q(>0)] ((x^(qk) (q+1)−kx^q (qk+1))/(k(q+1)(qk+1)))≥((1−k)/k) ((x^(qk) (q+1)−kx^q (qk+1))/((q+1)(qk+1)))≥1−k [Multiplying by q(>0)] x^(qk) (q+1)−kx^q (qk+1)≥(1−k)(q+1)(qk+1) qx^(ak) +x^(qk) −qk^2 x^q −kx^q ≥(1−k)(q^2 k+q+qk+1) qx^(ak) +x^(qk) −qk^2 x^q −kx^q ≥q^2 k+q+qk^(×) +1−q^2 k^2 −qk^(×) −qk^2 −k qx^(ak) +x^(qk) −qk^2 x^q −kx^q ≥q^2 k+q+1−q^2 k^2 −qk^2 −k

\frac{1}{p} (\frac{x^{p}}{p + 1} - 1) ⩾ \frac{1}{q} (\frac{x^{q}}{q + 1} - 1)

\frac{x^{p}}{p (p + 1)} - \frac{1}{p} ⩾ \frac{x^{q}}{q (q + 1)} - \frac{1}{q}

\frac{x^{p}}{p (p + 1)} - \frac{x^{q}}{q (q + 1)} ⩾ \frac{1}{p} - \frac{1}{q}

L e t p = q k, w h e r e k > 1

\frac{x^{q k}}{q k (q k + 1)} - \frac{x^{q}}{q (q + 1)} ⩾ \frac{1}{q k} - \frac{1}{q}

\frac{x^{q k}}{k (q k + 1)} - \frac{x^{q}}{(q + 1)} ⩾ \frac{1}{k} - 1 [M u l t k p l y i n g b y q (> 0)]

\frac{x^{q k} (q + 1) - {k x}^{q} (q k + 1)}{k (q + 1) (q k + 1)} ⩾ \frac{1 - k}{k}

\frac{x^{q k} (q + 1) - {k x}^{q} (q k + 1)}{(q + 1) (q k + 1)} ⩾ 1 - k [M u l t i p l y i n g b y q (> 0)]

x^{q k} (q + 1) - {k x}^{q} (q k + 1) ⩾ (1 - k) (q + 1) (q k + 1)

{q x}^{a k} + x^{q k} - {q k}^{2} x^{q} - {k x}^{q} ⩾ (1 - k) (q^{2} k + q + q k + 1)

{q x}^{a k} + x^{q k} - {q k}^{2} x^{q} - {k x}^{q} ⩾ q^{2} k + q + \overset{\times}{q k} + 1 - q^{2} k^{2} - \overset{\times}{q k} - {q k}^{2} - k

{q x}^{a k} + x^{q k} - {q k}^{2} x^{q} - {k x}^{q} ⩾ q^{2} k + q + 1 - q^{2} k^{2} - {q k}^{2} - k

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