A question on inequality and differentiation of logarithms

Question

Show by differentiating that $\ln x$ is a concave function of $x$. Deduce that if $p,q,x,y$ are positive real numbers with ${1\over p}+{1\over q}=1$, then $$xy \lt {x^p\over p}+{y^q\over q}$$

I can see why $f(x)=\ln x$ is concave by differentiating it twice giving $-{1\over x^2} \lt 0$, but for the later part of the question I have no clue.

How will you do this question? Please share your thought process, or indicate that it's your experience of solving numerous problems that give you the direction.

Answer 1

Note that $$\ln(x y)= \frac{1}{p} \ln x^p + \frac{1}{q} \ln y^q ,$$ Since $q^{-1}+p^{-1}=1$, we can use Jensen's inequality to show that $$\ln(x y)=\frac{1}{p} \ln x^p + \frac{1}{q} \ln y^q \leq \ln \left(\frac{x^{p}}{p} +\frac{y^q}{q} \right)$$ Now, just take the exponential to get the answer. not that the inequality is not strict (take for example $p=q=2$ and $x=y=1$.

Answer 2

As $\log$ is concave, you have, for all $u,v>0$ and for all $\lambda \in (0,1)$, $\log(\lambda u+(1-\lambda)v)>\lambda\log(u)+(1-\lambda)\log(v)$

Now try to apply this inequality to your problem.