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


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

$P r o v e t h a t, i f p > q > 0 a n d x ⩾ 0, t h e n$ $\frac{1}{p} (\frac{x^{p}}{p + 1} - 1) ⩾ \frac{1}{q} (\frac{x^{q}}{q + 1} - 1) .$
Commented by Rasheed Soomro last updated on 24/Oct/15

$P r o v e t h a t, i f p > q > 0 a n d x ⩾ 0, t h e n$ $\frac{1}{p} (\frac{x^{p}}{p + 1} - 1) ⩾ \frac{1}{q} (\frac{x^{q}}{q + 1} - 1) .$ $- - - - - - - - - - - - - - - -$ $A T r y . N o t m u c h c o n f i d e n t$ ${L e t}^{'} s t r y t o a c h i e v e s i m p l e r e q u i v a l e n t o f$ $t h e g i v e n s t a t m e n t, w h i c h i s e a s y t o p r o v e .$ $L e t p = q + k, w h e r e k > 0$ $\frac{1}{q + k} (\frac{x^{q + k}}{q + k + 1} - 1) ⩾ \frac{1}{q} (\frac{x^{q}}{q + 1} - 1)$ $\frac{1}{q + k} (\frac{x^{q + k} - q - k - 1}{q + k + 1}) ⩾ \frac{1}{q} (\frac{x^{q} - q - 1}{q + 1})$ $\frac{x^{q + k} - q - k - 1}{(q + k) (q + k + 1)} ⩾ \frac{x^{q} - q - 1}{q (q + 1)}$ $\frac{x^{q + k} - q - k - 1}{q^{2} + q + 2 q k + k^{2} + k} ⩾ \frac{x^{q} - q - 1}{q^{2} + q}$ $q > 0 \land k > 0 \Rightarrow 2 q k + k^{2} + k > 0$ $R e d u c t i o n o f 2 q k + k^{2} + k (a + v e n u m b e r)$ $i n t h e d e n o m i n a t o r o f g r e a t e r / e q u a l s i d e$ $w i l l l e a v e n o e f f e c t o n t h e i n e q u a l i t y .$ $H e n c e,$ $\frac{x^{q + k} - q - k - 1}{q^{2} + q} ⩾ \frac{x^{q} - q - 1}{q^{2} + q}$ $M u l t i p l y i n g b y q^{2} + q (> 0) t o b o t h s i d e s$ $x^{q + k} - q - k - 1 ⩾ x^{q} - q - 1$ $x^{q + k} - k ⩾ x^{q}$ $S i n c e k > 0$ $x^{q + k} - k + k ⩾ x^{q}$ $x^{q} . x^{k} ⩾ x^{q}$ $G i v e n t h a t x > 0 o r x = 0$ $f o r x > 0, x^{q} c a n b e c a n c e l l e d f r o m b o t h s i d e s :$ $x^{k} ⩾ 1$ ${L e t}^{'} s p r o v e f o r x ⩾ 1 f i r s t .$ $x^{k} ⩾ 1 i s c o m p l e t e l y e q u i v a l e n t o f$ $t h e g i v e n s t a t e m e n t a n d i t i s c l e a r l y t r u e s i n c e$ $x ⩾ 1 \land k > 0$ $H e n c e g i v e n s t a t e m e n t i s t r u e f o r x ⩾ 1$ $F o r x = 0 t h e r e s u l t c a n b e p r o v e d d i r e c t l y .$ $N o w t h e p r o b l e m i s f o r x \in (0, 1)$ $. . . .$ $C o n t i n u e$
Commented by prakash jain last updated on 24/Oct/15

$Observation on Rasheed Comment$ $\frac{x^{q + k} - q - k - 1}{q^{2} + q + 2 q k + k^{2} + k} ⩾ \frac{x^{2} - q - 1}{q^{2} + q}$ $is not equivalent to (if 2 q k + k^{2} + k > 0)$ $\frac{x^{q + k} - q - k - 1}{q^{2} + q} ⩾ \frac{x^{2} - q - 1}{q^{2} + q}$ $since \frac{x^{q + k} - q - k - 1}{q^{2} + q + 2 q k + k^{2} + k} < \frac{x^{q + k} - q - k - 1}{q^{2} + q}$ $Suggestion : inequality may be simpler to prove$ $if you try p = k q, k > 1$
Commented by Rasheed Soomro last updated on 24/Oct/15

$THANKS S S S S s s s . . . f o r c o r r e c t i o n a n d s u g g e s t i o n!$
Commented by Rasheed Soomro last updated on 25/Oct/15

$P r o v e t h a t, i f p > q > 0 a n d x ⩾ 0, t h e n$ $\frac{1}{p} (\frac{x^{p}}{p + 1} - 1) ⩾ \frac{1}{q} (\frac{x^{q}}{q + 1} - 1) .$ $- - - - - \times \times \times - - - - - - -$ $T o g e t a simpler equivalent (?) :$ $\frac{x^{p}}{p (p + 1)} - \frac{1}{p} ⩾ \frac{x^{q}}{q (q + 1)} - \frac{1}{q} . . . . . . . . . . . . . . . . . . . . . . (1)$ $N o w,$ $p > q \Rightarrow \frac{1}{p} < \frac{1}{q} \Rightarrow - \frac{1}{p} > - \frac{1}{q} . . . . . . . . . . (2)$ $A d d i n g (1) a n d (2)$ $\frac{x^{p}}{p (p + 1)} ⩾ \frac{x^{q}}{q (q + 1)}$ $I s t h i s e q u i v a l e n t t o t h e o r i g i n a l ?$ $D o e s a d d i n g a same - sense inequality y i e l d a n$ $equivalent inequality ?$ $L e t p = q k, w h e r e k > 1$ $\frac{x^{q k}}{q k (q k + 1)} ⩾ \frac{x^{q}}{q (q + 1)}$ $\frac{x^{q k}}{k (q k + 1)} ⩾ \frac{x^{q}}{q + 1} [M u l t i p l y i n g b y q (> 0)]$ $\frac{x^{q k}}{{q k}^{2} + k} ⩾ \frac{x^{q}}{q + 1}$ $Continue$
Commented by Rasheed Soomro last updated on 25/Oct/15

$On the comment of prakash jain$ $Sir y o u h a v e s a i d t h a t a > b i s n o t e q u i v a l e n t t o A > B$ $b e c a u s e (a f t e r d r o p p i n g s o m e + v e v a l u e f r o m d e n o m i n a t o r)$ $A < a .$ $D o e s t h i s m e a n a > b a n d A > B w i l l b e e q u i v a l e n t o n l y$ $w h e n A = a ?$
Commented by prakash jain last updated on 25/Oct/15

$While proving$ $if a < A then a > b \Rightarrow A > b but A > b ⇏ a > b$ $if a > A then A > b \Rightarrow a > b but a > b ⇏ A > b$
Commented by prakash jain last updated on 25/Oct/15

$Rasheed regarding equality for p > q$ $If \frac{x^{p}}{p (p + 1)} ⩾ \frac{x^{q}}{q (q + 1)} . . . . . . (1)$ $then \frac{x^{p}}{p (p + 1)} - \frac{1}{p} ⩾ \frac{x^{q}}{q (q + 1)} - \frac{1}{q} . . . . . . (2)$ $(1) \Rightarrow (2) however (2) does not imply (1) .$ $Now you cannot prove 1 since is not true .$ $Try x = 1, p = 2, q = 1$
Commented by Rasheed Soomro last updated on 25/Oct/15

$THANKS f o r s o m a n y v a l u e a b l e e x p l a n a t i o n s!$ $However I would like to understand clearly$ $why$ $\frac{x^{p}}{p (p + 1)} - \frac{1}{p} ⩾ \frac{x^{q}}{q (q + 1)} - \frac{1}{q}$ $does not \underline{imply} to$ $\frac{x^{p}}{p (p + 1)} ⩾ \frac{x^{q}}{q (q + 1)}$ $when p > q > 0 is also given ?$ $p > q \Rightarrow \frac{1}{p} < \frac{1}{q} \Rightarrow - \frac{1}{p} > - \frac{1}{q}$ $\frac{x^{p}}{p (p + 1)} - \frac{1}{p} ⩾ \frac{x^{q}}{q (q + 1)} - \frac{1}{q} \land - \frac{1}{p} > - \frac{1}{q}$ $\Rightarrow \frac{x^{p}}{p (p + 1)} ⩾ \frac{x^{q}}{q (q + 1)} [S u b t r a c t i n g s e c o n d i n e q u a l i t y f r o m f i r s t]$ $H o w e v e r a s f o r a s \underline{e q u i v a l e n c e} i s c o n c e r n e d I a d m i t$ $t h a t t h e s e t w o a r e n o t e q u i v a l e n t . P e r h a p s t h i s i s$ $t h e r e a s o n t h a t o n e s a t i s f y s o m e v a l u e s a n d o t h e r n o t .$ $B u t s i n c e t h e g o a l w a s t o d e t e r m i n e a n e q u i v a l e n t s o$ $I f a i l e d t o a c h i e v e .$ $I t h i n k i m p l i c a t i o n a n d e q u i v a l e n c e a r e t w o$ $d i f f e r e n t t h i n g s .$ $A n y w a y y o u r o p i n i o n w i l l b e a n {e x p e r t}^{'} s o p i n i o n$ $a n d m y q u e s t i o n i s a {s t u d e n t}^{'} s q u e z t i o n .$ $I a m h e r e t o l e a r n f r o m y o u a n d o t h e r e x p e r t s .$
Commented by Rasheed Soomro last updated on 27/Oct/15
$Prove that, if p>q>0 and x≥0, then (1/p)((x^p /(p+1))−1)≥(1/q)((x^q /(q+1))−1) −×−×−×−×−×−×−×−×−×− (x^p /(p(p+1)))−(1/p)≥(x^q /(q(q+1)))−(1/q) (x^p /(p(p+1)))−(x^q /(q(q+1)))≥(1/p)−(1/q) ((x^p q(q+1)−x^q p(p+1))/(pq(p+1)(q+1)))≥((q−p)/(pq)).....................(A) Approach(1) Let p=qk,where k>1_(−) ((x^p q(q+1)−x^q p(p+1))/(pq(p+1)(q+1)))≥((q−p)/(pq)).....................(A) ((x^(qk) q(q+1)−x^q qk(qk+1))/(q^2 k(qk+1)(q+1)))≥((q−qk)/(q^2 k)) ((x^(qk) q(q+1)−x^q qk(qk+1))/((qk+1)(q+1)))≥q(1−k) ((x^(qk) (q+1)−x^q k(qk+1))/((qk+1)(q+1)))≥1−k ((x^(qk) (q+1)−x^q k(qk+1))/((qk+1)(q+1)))+k≥1 ((x^(qk) (q+1)−x^q k(qk+1)+k(qk+1)(q+1))/((qk+1)(q+1))) ≥ 1 ((x^(qk) q+x^(qk) −x^q qk^2 −x^q k+q^2 k^2 +qk^2 +qk+k)/((qk+1)(q+1)))≥1 ((x^(qk) q+x^(qk) −x^q qk^2 −x^q k+q^2 k^2 +qk^2 +qk+k)/(q^2 k+qk+q+1))≥1 x^(qk) q+x^(qk) −x^q qk^2 −x^q k+q^2 k^2 +qk^2 +qk+k−q^2 k−qk−q−1≥0 ...... Approach(2) Let p=q+k,where k>0 ((x^p q(q+1)−x^q p(p+1))/(pq(p+1)(q+1)))≥((q−p)/(pq)).....................(A) ((x^((q+k)) q(q+1)−x^q (q+k){(q+k)+1})/(q(q+1){(q+k)+1}(q+1)))≥((q−(q+k))/(q(q+k))) ((x^(q+k) q(q+1)−x^q (q+k)(q+k+1))/(q(q+1)^2 (q+k+1)))≥((−k)/(q(q+k))) ((x^(q+k) q(q+1)−x^q (q+k)(q+k+1))/((q+1)^2 (q+k+1)))≥((−k)/(q+k)) [Multiply by q(>0)] ((x^(q+k) q(q+1)−x^q (q+k)(q+k+1))/((q+1)^2 (q+k+1)))+(k/(q+k)) ≥0 (({x^(q+k) q(q+1)−x^q (q+k)(q+k+1)}(q+k)+k(q+1)^2 (q+k+1))/((q+1)^2 (q+k)(q+k+1))) ≥0 Continue$
$P r o v e t h a t, i f p > q > 0 a n d x ⩾ 0, t h e n$ $\frac{1}{p} (\frac{x^{p}}{p + 1} - 1) ⩾ \frac{1}{q} (\frac{x^{q}}{q + 1} - 1)$ $- \times - \times - \times - \times - \times - \times - \times - \times - \times -$ $\frac{x^{p}}{p (p + 1)} - \frac{1}{p} ⩾ \frac{x^{q}}{q (q + 1)} - \frac{1}{q}$ $\frac{x^{p}}{p (p + 1)} - \frac{x^{q}}{q (q + 1)} ⩾ \frac{1}{p} - \frac{1}{q}$ $\frac{x^{p} q (q + 1) - x^{q} p (p + 1)}{p q (p + 1) (q + 1)} ⩾ \frac{q - p}{p q} . . . . . . . . . . . . . . . . . . . . . (A)$ $A p p r o a c h (1)$ $\underline{L e t p = q k, w h e r e k > 1}$ $\frac{x^{p} q (q + 1) - x^{q} p (p + 1)}{p q (p + 1) (q + 1)} ⩾ \frac{q - p}{p q} . . . . . . . . . . . . . . . . . . . . . (A)$ $\frac{x^{q k} q (q + 1) - x^{q} q k (q k + 1)}{q^{2} k (q k + 1) (q + 1)} ⩾ \frac{q - q k}{q^{2} k}$ $\frac{x^{q k} q (q + 1) - x^{q} q k (q k + 1)}{(q k + 1) (q + 1)} ⩾ q (1 - k)$ $\frac{x^{q k} (q + 1) - x^{q} k (q k + 1)}{(q k + 1) (q + 1)} ⩾ 1 - k$ $\frac{x^{q k} (q + 1) - x^{q} k (q k + 1)}{(q k + 1) (q + 1)} + k ⩾ 1$ $\frac{x^{q k} (q + 1) - x^{q} k (q k + 1) + k (q k + 1) (q + 1)}{(q k + 1) (q + 1)} ⩾ 1$ $\frac{x^{q k} q + x^{q k} - x^{q} {q k}^{2} - x^{q} k + q^{2} k^{2} + {q k}^{2} + q k + k}{(q k + 1) (q + 1)} ⩾ 1$ $\frac{x^{q k} q + x^{q k} - x^{q} {q k}^{2} - x^{q} k + q^{2} k^{2} + {q k}^{2} + q k + k}{q^{2} k + q k + q + 1} ⩾ 1$ $x^{q k} q + x^{q k} - x^{q} {q k}^{2} - x^{q} k + q^{2} k^{2} + {q k}^{2} + q k + k - q^{2} k - q k - q - 1 ⩾ 0$ $. . . . . .$ $A p p r o a c h (2)$ $L e t p = q + k, w h e r e k > 0$ $\frac{x^{p} q (q + 1) - x^{q} p (p + 1)}{p q (p + 1) (q + 1)} ⩾ \frac{q - p}{p q} . . . . . . . . . . . . . . . . . . . . . (A)$ $\frac{x^{(q + k)} q (q + 1) - x^{q} (q + k) {(q + k) + 1}}{q (q + 1) {(q + k) + 1} (q + 1)} ⩾ \frac{q - (q + k)}{q (q + k)}$ $\frac{x^{q + k} q (q + 1) - x^{q} (q + k) (q + k + 1)}{q {(q + 1)}^{2} (q + k + 1)} ⩾ \frac{- k}{q (q + k)}$ $\frac{x^{q + k} q (q + 1) - x^{q} (q + k) (q + k + 1)}{{(q + 1)}^{2} (q + k + 1)} ⩾ \frac{- k}{q + k} [M u l t i p l y b y q (> 0)]$ $\frac{x^{q + k} q (q + 1) - x^{q} (q + k) (q + k + 1)}{{(q + 1)}^{2} (q + k + 1)} + \frac{k}{q + k} ⩾ 0$ $\frac{{x^{q + k} q (q + 1) - x^{q} (q + k) (q + k + 1)} (q + k) + k {(q + 1)}^{2} (q + k + 1)}{{(q + 1)}^{2} (q + k) (q + k + 1)} ⩾ 0$ $C o n t i n u e$
Commented by Rasheed Soomro last updated on 26/Oct/15
$(1/p)((x^p /(p+1))−1)≥(1/q)((x^q /(q+1))−1) p>0,q>0⇒pq>0,multiplying by pq to both sides q((x^p /(p+1))−1)≥p((x^q /(q+1))−1) ((qx^p )/(p+1))−q≥((px^q )/(q+1))−p_ ((qx^p )/(p+1))+p≥((px^q )/(q+1))+q [Adding p+q to both sides] ((qx^p +p(p+1))/(p+1))≥((px^q +q(q+1))/(q+1)) Let p=qk, where k>1 ((qx^(qk) +qk(qk+1))/(qk+1))≥((qkx^q +q(q+1))/(q+1)) q>0 ∧ k>0⇒q+1>0 ∧ qk+1>0 ⇒(qk+1)(q+1)>0 Multiplying by (qk+1)(q+1) to both sides: q{x^(qk) +qk^2 +k}(q+1)≥q{kx^q +q+1}(qk+1) }÷q (>0) {x^(qk) +qk^2 +k}(q+1)≥{kx^q +q+1}(qk+1) qx^(qk) +q^2 k^2 +qk^(×) +x^(qk) +qk^2 +k≥qk^2 x^q +q^2 k+qk^(×) +kx^q +q+1 qx^(qk) +q^2 k^2 +x^(qk) +qk^2 +k≥qk^2 x^q +q^2 k+kx^q +q+1 (q+1)x^(qk) +q^2 k^2 +qk^2 +k≥k(qk+1)x^q +q^2 k+q+1 The above inequality is equivalent to the inequality to be proved.To prove the given it is suficient to prove its equivalent. If we could prove the following inequalities They imply aove inequality. Howdver vice versa is not correct. (q+1)x^(qk) ≥k(qk+1)x^q [.....].....................(i) q^2 k^2 ≥q^2 k⇒(q^2 k)k≥q^2 k [∵ k>1]..........(ii) qk^2 ≥q [k>1 ∧ q>0⇒k^2 >1⇒qk^2 ≥q.....(iii) k≥1 [Assumption]...........................(iv) Unfortinuately (i) is not always correct.For example at x=1 it is false! Dilli hunooz door ast Continue$
$\frac{1}{p} (\frac{x^{p}}{p + 1} - 1) ⩾ \frac{1}{q} (\frac{x^{q}}{q + 1} - 1)$ $p > 0, q > 0 \Rightarrow p q > 0, m u l t i p l y i n g b y p q t o b o t h s i d e s$ $q (\frac{x^{p}}{p + 1} - 1) ⩾ p (\frac{x^{q}}{q + 1} - 1)$ $\frac{{q x}^{p}}{p + 1} - q ⩾ \frac{{p x}^{q}}{q + 1} - p_{}$ $\frac{{q x}^{p}}{p + 1} + p ⩾ \frac{{p x}^{q}}{q + 1} + q [A d d i n g p + q t o b o t h s i d e s]$ $\frac{{q x}^{p} + p (p + 1)}{p + 1} ⩾ \frac{{p x}^{q} + q (q + 1)}{q + 1}$ $L e t p = q k, w h e r e k > 1$ $\frac{{q x}^{q k} + q k (q k + 1)}{q k + 1} ⩾ \frac{{q k x}^{q} + q (q + 1)}{q + 1}$ $q > 0 \land k > 0 \Rightarrow q + 1 > 0 \land q k + 1 > 0 \Rightarrow (q k + 1) (q + 1) > 0$ $M u l t i p l y i n g b y (q k + 1) (q + 1) t o b o t h s i d e s :$ $q {x^{q k} + {q k}^{2} + k} (q + 1) ⩾ q {{k x}^{q} + q + 1} (q k + 1)} \div q (> 0)$ ${x^{q k} + {q k}^{2} + k} (q + 1) ⩾ {{k x}^{q} + q + 1} (q k + 1)$ ${q x}^{q k} + q^{2} k^{2} + \overset{\times}{q k} + x^{q k} + {q k}^{2} + k ⩾ {q k}^{2} x^{q} + q^{2} k + \overset{\times}{q k} + {k x}^{q} + q + 1$ ${q x}^{q k} + q^{2} k^{2} + x^{q k} + {q k}^{2} + k ⩾ {q k}^{2} x^{q} + q^{2} k + {k x}^{q} + q + 1$ $(q + 1) x^{q k} + q^{2} k^{2} + {q k}^{2} + k ⩾ k (q k + 1) x^{q} + q^{2} k + q + 1$ $T h e a b o v e i n e q u a l i t y i s e q u i v a l e n t t o t h e i n e q u a l i t y$ $t o b e p r o v e d . T o p r o v e t h e g i v e n i t i s s u f i c i e n t t o$ $p r o v e i t s e q u i v a l e n t .$ $I f w e c o u l d p r o v e t h e f o l l o w i n g i n e q u a l i t i e s$ $T h e y i m p l y a o v e i n e q u a l i t y . H o w d v e r v i c e v e r s a$ $i s n o t c o r r e c t .$ $(q + 1) x^{q k} ⩾ k (q k + 1) x^{q} [. . . . .] . . . . . . . . . . . . . . . . . . . . . (i)$ $q^{2} k^{2} ⩾ q^{2} k \Rightarrow (q^{2} k) k ⩾ q^{2} k [∵ k > 1] . . . . . . . . . . (i i)$ ${q k}^{2} ⩾ q [k > 1 \land q > 0 \Rightarrow k^{2} > 1 \Rightarrow {q k}^{2} ⩾ q . . . . . (i i i)$ $k ⩾ 1 [A s s u m p t i o n] . . . . . . . . . . . . . . . . . . . . . . . . . . . (i v)$ $U n f o r t i n u a t e l y (i) i s n o t a l w a y s c o r r e c t . F o r e x a m p l e$ $a t x = 1 i t i s f a l s e!$ $D i l l i h u n o o z d o o r a s t$ $C o n t i n u e$
	Answered by Rasheed Soomro last updated on 26/Oct/15

	$E x p e r i m e n t$ $\frac{1}{p} (\frac{x^{p}}{p + 1} - 1) ⩾ \frac{1}{q} (\frac{x^{q}}{q + 1} - 1)$ $L e t p = q^{k}, w h e r e k > 1$ $\frac{1}{q^{k}} (\frac{x^{q^{k}}}{q^{k} + 1} - 1) ⩾ \frac{1}{q} (\frac{x^{q}}{q + 1} - 1)$ $\frac{1}{q^{k - 1}} (\frac{x^{q^{k}}}{q^{k} + 1} - 1) ⩾ (\frac{x^{q}}{q + 1} - 1)$ $\frac{x^{q^{k}}}{q^{q - 1} (q^{k} + 1)} - \frac{1}{q^{k - 1}} ⩾ \frac{x^{q} - q - 1}{q + 1}$ $\frac{x^{q^{k}} - q^{k} - 1}{q^{k - 1} (q^{k} + 1)} ⩾ \frac{x^{q} - q - 1}{q + 1}$ $(x^{q^{k}} - q^{k} - 1) (q + 1) ⩾ q^{k - 1} (q^{k} + 1) (x^{q} - q - 1)$ ${q x}^{q^{k}} - q^{k + 1} - q + x^{q^{k}} - q^{k} - 1 ⩾ q^{k - 1} (q^{k} x^{q} - q^{k + 1} - q^{k} + x^{q} - q - 1)$ ${q x}^{q^{k}} - q^{k + 1} - q + x^{q^{k}} - q^{k} - 1 ⩾ q^{2 k - 1} x^{q} - q^{2 k} - q^{2 k - 1} + q^{k - 1} x^{q} - q^{k} - q^{k - 1})$ $(q + 1) x^{q^{k}} - q^{k + 1} - q - q^{k} - 1 ⩾ (q^{2 k - 1} + q^{k - 1}) x^{q} - q^{2 k} - q^{2 k - 1} - q^{k} - q^{k - 1})$
	Answered by Rasheed Soomro last updated on 28/Oct/15

	$\frac{1}{p} (\frac{x^{p}}{p + 1} - 1) ⩾ \frac{1}{q} (\frac{x^{q}}{q + 1} - 1)$ $\frac{x^{p}}{p (p + 1)} - \frac{1}{p} ⩾ \frac{x^{q}}{q (q + 1)} - \frac{1}{q}$ $\frac{x^{p}}{p (p + 1)} - \frac{x^{q}}{q (q + 1)} ⩾ \frac{1}{p} - \frac{1}{q}$ $L e t p = q k, w h e r e k > 1$ $\frac{x^{q k}}{q k (q k + 1)} - \frac{x^{q}}{q (q + 1)} ⩾ \frac{1}{q k} - \frac{1}{q}$ $\frac{x^{q k}}{k (q k + 1)} - \frac{x^{q}}{(q + 1)} ⩾ \frac{1}{k} - 1 [M u l t k p l y i n g b y q (> 0)]$ $\frac{x^{q k} (q + 1) - {k x}^{q} (q k + 1)}{k (q + 1) (q k + 1)} ⩾ \frac{1 - k}{k}$ $\frac{x^{q k} (q + 1) - {k x}^{q} (q k + 1)}{(q + 1) (q k + 1)} ⩾ 1 - k [M u l t i p l y i n g b y q (> 0)]$ $x^{q k} (q + 1) - {k x}^{q} (q k + 1) ⩾ (1 - k) (q + 1) (q k + 1)$ ${q x}^{a k} + x^{q k} - {q k}^{2} x^{q} - {k x}^{q} ⩾ (1 - k) (q^{2} k + q + q k + 1)$ ${q x}^{a k} + x^{q k} - {q k}^{2} x^{q} - {k x}^{q} ⩾ q^{2} k + q + \overset{\times}{q k} + 1 - q^{2} k^{2} - \overset{\times}{q k} - {q k}^{2} - k$ ${q x}^{a k} + x^{q k} - {q k}^{2} x^{q} - {k x}^{q} ⩾ q^{2} k + q + 1 - q^{2} k^{2} - {q k}^{2} - k$
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