Working on a question Proof that if $x,y>0$ and $x+y=1$, then $(2x)^{\frac 1 x}+(2y)^{\frac 1 y}\leq 2$ . I find a connection between Bernoulli's inequality and Young's inequality :

Let $0< r\leq 1$ and $x\geq0$ then prove that : $$(1+x)^r\leq 1+rx$$

We start with the LHS and apply Young's inequality :

$$1(1+x)^r\leq \frac{(1+x)^{rp}}{p}+\frac{1}{q}$$

With $\frac{1}{q}+\frac{1}{p}=1$ with $p,q>0$

we put now $p=\frac{1}{r}$ to get :

$$(1+x)^r\leq (1+x)r+1-r$$

Wich is the desired inequality .


Is it fine ? Can we improve the situation ?

Thanks in advance !


  • $\begingroup$ Your proof looks correct to me. However, it is unclear to me what you mean with “improving the situation”. $\endgroup$ – Martin R Aug 23 at 3:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.