Proving inequality with derivatives

Question

In my course we have following exercise:

Let $p>0, q>0, 1/p + 1/q =1$.

Prove that: $$ xy \leq \frac{1}p x^p+ \frac{1}{q} y^q \quad(x>0,y>0) $$

Protip: calculate maximum value of function $$\displaystyle t \rightarrow t^ \frac{1}{p}- \frac{t}{p} \quad (t>0) $$

I have calculated protip function maximum but don't know where to go from there.

Answer 1

Let $x=u^{1/p}$ and $y=v^{1/q}$. The inequality to prove becomes

$$u^{1/p}v^{1/q}\le{1\over p}u+{1\over q}v$$

Writing $u=tv$ with $0\lt t$, we find that the inequality to prove is

$$t^{1/p}v^{{1\over p}+{1\over q}}\le\left({1\over p}t+{1\over q}\right)v$$

But since ${1\over p}+{1\over q}=1$, we have a (positive) $v$ on both sides. After cancelling it, all we need to prove is

$$t^{1/p}-{1\over p}t\le{1\over q}=1-{1\over p}$$

so it suffices to show that the function $t\to t^{1/p}-{t\over p}$ attains its maximum value at $t=1$. (Note, we must have $p\gt1$ in order to have ${1\over p}+{1\over q}=1$ with $q\gt0$.)

Answer 2

It is easy with Jensen's inequality:

Compare the logs. Since $\ln$ is a concave function, Jensen's inequality ensures that $$\ln\biggl(\frac{1}p x^p+ \frac{1}{q} y^q\biggr) \ge\frac 1p\ln(x^p)+\frac1q\ln(y^q)$$

The right hand side is equal to:

$$\frac 1p\ln(x^p)+\frac1q\ln(y^q)=\ln x+\ln y=\ln(xy),$$ whence the requested inequality.