Prove $xy\leq \frac{x^p}{p} + \frac{y^q}{q}$

Question

Let $p,q>1,\ \frac1p+\frac1q=1$, and $x,y>0$.

Prove that $xy\leq \frac{x^p}{p} + \frac{y^q}{q}$ by using natural log, definition of concave function, and the fact that natural log is a concave function

Concave function: $f(tx+(1-t)y)\geq tf(x)+(1-t)f(y)$
for every $x,y$ in its domain and $0\leq t \leq 1$

Answer 1

I think what you really want to do is to prove the inequality for $x,y>0$ not that the inequality implies $x,y>0$ since it, for example, also holds for $x=y=-2,\ p=q=2$.

So how do we prove this inequality given the convexity of $\ln$ and $x,y>0$? By using convexity, logarithm laws and our previously stated conditions we get

$$\ln\Big( \frac1p x^p + \frac1q y^q \Big) = \ln\Big( \frac1p x^p + \big(1-\frac1p \big) y^q \Big)$$ $$\geq \frac1p \ln(x^p) + \big( 1-\frac1p\big) \ln(y^q) = \frac1p \ln(x^p) + \frac1q \ln(y^q) = \ln(x) + \ln(y)$$

Appliying $\exp$ on both sides gives you:

$$\frac1p x^p + \frac1q y^q \geq xy$$