Question Number 1952 by prakash jain last updated on 25/Oct/15
$$\mathrm{Inequality}\:\mathrm{relation}\:\mathrm{starting}\:\mathrm{a}\:\mathrm{new}\:\mathrm{thread} \\ $$$$\frac{{x}^{{p}} }{{p}\left({p}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{p}}\geqslant\frac{{x}^{{q}} }{{q}\left({q}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{q}} \\ $$$${p}=\mathrm{2},\:{q}=\mathrm{1},\:{x}=\mathrm{1} \\ $$$$\frac{{x}^{{p}} }{{p}\left({p}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\frac{{x}^{{q}} }{{q}\left({q}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\frac{{x}^{{p}} }{{p}\left({p}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{p}}\:=\:\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\frac{{x}^{{p}} }{{p}\left({p}+\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:−\:\mathrm{1}\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${A}\geqslant{B}\:\mathrm{and}\:{c}>{d}\:\nRightarrow{A}−{c}\geqslant{B}−{d} \\ $$$$\mathrm{Since}\:\mathrm{you}\:\mathrm{are}\:\mathrm{subtracting}\:\mathrm{a}\:\mathrm{larger}\:\mathrm{quantity} \\ $$$${c}\:\mathrm{from}\:{A},\:\mathrm{so}\:\mathrm{if}\:\left({c}−{d}\right)>\left({A}−{B}\right)\:\mathrm{then}\:{A}−{c}<{B}−{d} \\ $$
Commented by prakash jain last updated on 25/Oct/15
$$\mathrm{Please}\:\mathrm{see}\:\mathrm{comments}\:\mathrm{in}\:\mathrm{Q1928} \\ $$
Commented by Rasheed Soomro last updated on 26/Oct/15
$${I}\:{learnt}\:{something}\:{that}\:{I}\:{didn}'{t}\:{know}\:{before}! \\ $$