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

Question

Question:

Let $x \geq 0$ , $y \geq 0$ and $p > 0$, $q>0$ with $\frac{1}{p} + \frac{1}{q} = 1$.
Show that $$xy \leq \frac{x^p}{p} + \frac{y^q}{q} $$ [Suggestion: Without loss of generality suppose $xy = 1$].

Attempt: Let $f, \varphi : U \to \mathbb R$, $U = \{(x,y) \in \mathbb R^2; x > 0 , y > 0 \}$ given by $f(x,y) = \frac{x^p}{p} + \frac{y^q}{q}$ and $\varphi (x,y) = xy$. Then we have

$$\mathrm {grad}\, f(x,y) = (x^{p-1}, y^{q-1}) \,\,\,\text{and}\,\,\, \mathrm {grad} \,\varphi (x,y) = (y,x)$$

Then $1$ is a regular value of $\varphi$. Consider $M = \varphi^{-1} (1)$, the hyperbola $xy =1$. Now $(x,y) \in M$ is a critical point of $f|_M$ iff

$$\mathrm {grad}\, f(x,y) = \lambda\, \mathrm{grad} \, \varphi (x,y) \,\,\,\text{and}\,\,\, \varphi (x,y) = 1$$

As $x> 0 $ and $y>0$ we have

$$x^{p-1} = \lambda y \,\,\, , y ^{q-1} = \lambda x \,\,\,\text{and} \,\,\,xy = 1$$

Then $$\frac{x}{y} = \frac{y ^{q-1}}{x ^{p-1}} \implies x^p = y^q$$

This gives us $$\begin{align}\frac{x^p}{p} + \frac{y^q}{q} &= \frac{qy^q + py^q}{pq } = y^q \frac{p + q}{pq}\\&= y^q = y ^{1 + \frac{q}{p}}\\&=y^{\frac{q}{p}}\cdot y = x \cdot y\end{align}$$

Now $f$ is of class $C^{\infty}$ and its Hessian is given by

$$Hf(x,y) = \begin{pmatrix} (p-1)x^{p-2} & 0 \\ 0 & (q-1)y^{q-2}\end{pmatrix} $$

and it is positive, therefore $xy$ is a local minimum. It follows then $$xy \leq \frac{x^p}{p} + \frac{y^q}{q}$$

as we wanted.

The cases $x = y = 0$, $x = 0 $ and $y> 0$ were considered trivially true.

Note: This inequality is used to prove Hölder's Inequality.

Answer 1

Your proof is fine, up to a little typo: It should be $M = \varphi^{-1}(1)$ instead of $M = \varphi(1)$.

Answer 2

Use Jensen's inequality. The function $f(x)=-\log x$ is convex, and we have $\frac1p +\frac1q=1$, so $$f\left(\frac1p x^p +\frac 1q y^q\right)\le \frac1p f(x^p)+\frac1q f(y^q).$$ Plugging in $f$, this simplifies to $$-\log\left(\frac{x^p}p +\frac {y^q}q\right)\le -\log(xy).$$