Question Number 144095 by mathdanisur last updated on 21/Jun/21
$${if}\:\:{x}>\mathrm{0}\:,\:{r}={pq}\:,\:\mathrm{1}\leqslant{p}\leqslant{q}\:\:{then}: \\ $$$$\mathrm{1}+{rx}\:\leqslant\:\left(\mathrm{1}+{qx}\right)^{\boldsymbol{{p}}} \:\leqslant\:\left(\mathrm{1}+{px}\right)^{\boldsymbol{{q}}} \:\leqslant\:\left(\mathrm{1}+{x}\right)^{\boldsymbol{{r}}} \\ $$
Answered by mitica last updated on 21/Jun/21
![(1+x)^r =[(1+x)^p ]^q ≥^(bernoulli) (1+px)^q (1+qx)^p ≥^(bernoulli) 1+pqx=1+rx f(t)=((ln(1+t))/t) ↘ px≤qx⇒((ln(1+px))/p)≥((ln(1+qx))/q)⇒(1+px)^q ≥(1+qx)^p](https://www.tinkutara.com/question/Q144103.png)
$$\left(\mathrm{1}+{x}\right)^{{r}} =\left[\left(\mathrm{1}+{x}\right)^{{p}} \right]^{{q}} \:\overset{{bernoulli}} {\geqslant}\left(\mathrm{1}+{px}\right)^{{q}} \\ $$$$\left(\mathrm{1}+{qx}\right)^{{p}} \:\overset{{bernoulli}} {\geqslant}\mathrm{1}+{pqx}=\mathrm{1}+{rx} \\ $$$${f}\left({t}\right)=\frac{{ln}\left(\mathrm{1}+{t}\right)}{{t}}\:\searrow \\ $$$${px}\leqslant{qx}\Rightarrow\frac{{ln}\left(\mathrm{1}+{px}\right)}{{p}}\geqslant\frac{{ln}\left(\mathrm{1}+{qx}\right)}{{q}}\Rightarrow\left(\mathrm{1}+{px}\right)^{{q}} \geqslant\left(\mathrm{1}+{qx}\right)^{{p}} \\ $$
Commented by mathdanisur last updated on 21/Jun/21
$${alot}\:{perfect}\:{Sir}\:{thank}\:{you} \\ $$