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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.

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  • $\begingroup$ You can see the proof in Theorem 1.2.3 in the book Functional analysis. An introduction by Eidelman, Milman, Tsolomitis. $\endgroup$
    – Svetoslav
    Commented Mar 4, 2016 at 17:11
  • $\begingroup$ Using Multipliers? Because I know the proof using Mean Value Theorem. $\endgroup$ Commented Mar 4, 2016 at 17:13
  • $\begingroup$ You could also prove this result by using the concavity of $\log$. $\endgroup$
    – Augustin
    Commented Mar 4, 2016 at 17:14
  • $\begingroup$ I'm studying Analysis on $\mathbb R^n$, it's an exercise. $\endgroup$ Commented Mar 4, 2016 at 17:14
  • 1
    $\begingroup$ It is actually a geometric reasoning. $\endgroup$
    – Svetoslav
    Commented Mar 4, 2016 at 17:15

3 Answers 3

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

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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 $$-\log\left(\frac{x^p}p +\frac {y^q}q\right)\le -\log(xy).$$

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  • $\begingroup$ This is not an answer. The OP ask whether his/her attempt of proof is valid. $\endgroup$
    – gerw
    Commented Mar 4, 2016 at 18:35
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While this response may not directly address the posed question, it aims to provide a rationale for assuming $xy=1$ without loss of generality, as raised in the comments by @Bruno Mazeto.

The slight change that I want to note is as follows: Given $x^p=y^q$ and $xy=1$, we can observe that $$x^p=y^q\implies x^py^{-q}=1\implies x^{p+q}=1.$$ This implies $x=y=1$. Following the same steps as OP did we can conclude $$\dfrac{x^p}{p}+\dfrac{y^q}{q}\geq\frac1p+\frac1q,\tag{1}$$ where $p,q>1$ and $x,y>0$ are such $xy=1$.

Now, let's assume that $\dfrac{1}{p} + \dfrac{1}{q} = 1$ and consider the values $x = \dfrac{a}{(ab)^{1/p}}$ and $y = \dfrac{b}{(ab)^{1/q}}$, where $a, b > 0$. It is evident that $xy = 1$. Substituting these into equation (1) yields:

\begin{align} \dfrac{1}{p} \cdot \dfrac{a^p}{ab} + \dfrac{1}{q} \cdot \dfrac{b^q}{ab} \geq \dfrac{1}{p} + \dfrac{1}{q} = 1 \implies \dfrac{a^p}{p} + \dfrac{b^q}{q} \geq ab. \end{align}

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