Young's and Peter-Paul's inequalities

Question

Following the idea from Jarchow 1981, pp. 47–55, let's retell the whole story in case some wishes to be complete.

Let $1<p<\infty$ and $q$ such that $p+q=pq$, or \begin{align} \frac{1}{p}+\frac{1}{q}=1 \end{align} Let a real-valued function $f$ of the positive real number $t\ge0$ be defined as \begin{align} f(t):=\frac{t^{p} a^{p}}{p}+\frac{t^{-q}b^{q}}{q}\tag{1} \end{align} By setting the derivative over $t$ equal to zero, it can be found that \begin{align} t=(a^{-p}b^q)^{\frac{1}{p+q}} \end{align} Plugging it into (1), then the minimum of the function $f(t)$ can be found \begin{align} f(t)&=\frac{t^{p}a^{p}}{p}+\frac{t^{-q}b^{q}}{q}\\ &\ge\frac{(a^{-p}b^q)^{\frac{p}{p+q}}a^{p}}{p}+\frac{(a^{-p}b^q)^{\frac{-q}{p+q}}b^{q}}{q}\\ &=\frac{a^{\frac{pq}{p+q}}b}{p}+\frac{ab^{\frac{pq}{p+q}}}{q}\\ &=ab(\frac{1}{p}+\frac{1}{q})=ab. \end{align}

This is all right until I need to find what relation can be built between $t$ and $\varepsilon$ in the so-called "Peter–Paul" inequality \begin{align} ab\le \frac{a^2}{2\varepsilon}+\frac{\varepsilon b^2}{2} \end{align} since I will not be able to cancel the auxiliary parameter $t$ after using AM-GM. What on earth $p$ must be equal to $q$?

Answer 1

Fixing $a,b$, wat you have shown is that, for any conjugate pair $p,q>1$, and every $t\geq 0$, you have $$ ab \leq f(t) = \frac{t^pa^p}{p}+\frac{t^{-q}b^q}{q} \tag{1} $$ Now, given $\varepsilon>0$, you have to choose $p,q,t$ to get the Peter–Paul inequality. Since $a,b$ both have exponent $2$ there, it's natural to try $p=q=2$ (note that this is indeed a conjugate pair, as $1/2+1/2=1$!). This gives $$ ab \leq \frac{t^2a^2}{2}+\frac{b^2}{2t^2} \tag{2} $$ At this point, your hands are tied: you only parameter left is $t\geq 0$, and comparing what you have (e.g., the first term, $\frac{t^2a^2}{2}$) with what you want ($\frac{a^2}{2\varepsilon}$), you should choose $t = 1/\sqrt{\varepsilon}$. This give $$ ab \leq \frac{a^2}{2\varepsilon}+\frac{\varepsilon b^2}{2} \tag{3} $$ which is what you wanted.