# Young's and Peter-Paul's inequalities

Following the idea from Jarchow 1981, pp. 47–55, let's retell the whole story in case some wishes to be complete.
Let $$1 and $$q$$ such that $$p+q=pq$$, or $$\frac{1}{p}+\frac{1}{q}=1$$ Let a real-valued function $$f$$ of the positive real number $$t>0$$ be defined as $$f(t):=\frac{t^{p} a^{p}}{p}+\frac{t^{-q}b^{q}}{q}\label{1}\tag{1}$$ By setting the derivative over $$t$$ equal to zero, it can be found that $$t=(a^{-p}b^q)^{\frac{1}{p+q}}$$ Plugging it into \eqref{1}, then the minimum of the function $$f(t)$$ can be found \begin{align} f(t)&=\frac{t^{p}a^{p}}{p}+\frac{t^{-q}b^{q}}{q}\\ &\ge\frac{(a^{-p}b^q)^{\frac{p}{p+q}}a^{p}}{p}+\frac{(a^{-p}b^q)^{\frac{-q}{p+q}}b^{q}}{q}\\ &=\frac{a^{\frac{pq}{p+q}}b}{p}+\frac{ab^{\frac{pq}{p+q}}}{q}=ab\left(\frac{1}{p}+\frac{1}{q}\right)=ab. \end{align} This is all right until I need to find what relation can be built between $$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$$?

• In the inequality $f(t) \ge ab$ you could set $t:=\epsilon^{1/p}$. For $p=q=2$ you would recover the PP-inequality.
– daw
Commented May 16 at 9:46

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.
• It would be trivial when $p=q$. But, what I am after is the relation between $t$ and $\varepsilon$ for the case $p\neq q$. Commented Feb 21, 2022 at 10:23
• @MathArt I guess I don't understand your question then. What are you trying to prove? If it's the Peter-Paul inequality, then why don't you want to take $p=q=2$? Commented Feb 21, 2022 at 11:32
• Thanks. Now it seems that I have to be satisfied with $p=q=2$. Commented Apr 24, 2022 at 19:51