Inequality-relation-starting-a-new-thread-x-p-p-p-1-1-p-x-q-q-q-1-1-q-p-2-q-1-x-1-x-p-p-p-1-1-6-x-q-q-q-1-1-2-x-p-p-p-1-1-p-1-6-1-2-1-3-x-p-p-p-

Question Number 1952 by prakash jain last updated on 25/Oct/15

Inequality relation starting a new thread

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

p = 2, q = 1, x = 1

\frac{x^{p}}{p (p + 1)} = \frac{1}{6}

\frac{x^{q}}{q (q + 1)} = \frac{1}{2}

\frac{x^{p}}{p (p + 1)} - \frac{1}{p} = \frac{1}{6} - \frac{1}{2} = - \frac{1}{3}

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

A ⩾ B and c > d ⇏ A - c ⩾ B - d

Since you are subtracting a larger quantity

c from A, so if (c - d) > (A - B) then A - c < B - d

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

Please see comments in Q 1928

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

I l e a r n t s o m e t h i n g t h a t I {d i d n}^{'} t k n o w b e f o r e!