# A proof of Young's inequality

I need to prove that if $1 < p < \infty$ and $a, b \geqslant 0$ then $$ab \leqslant \frac{a^{p}}{p} + \frac{b^{q}}{q}$$ where $\frac 1p+\frac 1q=1$.

I fix $b$ and maximize the function $f(a) = ab - \frac{a^{p}}{p}$, but the maximum I find is $b^{q}$ with $q = \frac{1}{p-1}$. I have no idea how to get $q$ in the denominator. Any suggestion?

• I believe some hypotheses are missing... see Young's inequality en.wikipedia.org/wiki/Young%27s_inequality Commented Nov 14, 2013 at 1:31
• As it stands this cannot be true. Take $a=b=\frac{1}{2}$. Then let $p=q=100$. Commented Nov 14, 2013 at 1:33
• Weird. My book has it just as I posted it. Thanks for the help. Commented Nov 14, 2013 at 1:34
• The statement of the inequality is incomplete. Note that $p$ and $q$ must satisfy the condition $\frac{1}{p}+\frac{1}{q}=1$. Commented Nov 14, 2013 at 1:51
• Commented Jun 4, 2017 at 9:49

Let's prove that $f(x)=\frac {x^p}p+\frac{y^q}q-xy \geqslant 0$.

I'm going to calculate its minima for $x,y>0.$

$$f'(x)=x^{p-1}-y=0$$

So

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

since $(p-1)(q-1)=1$ by assumption.

Hence

$$f(x) \geqslant \frac{y^{p(q-1)}}p+\frac{y^q}q-y^{(q-1)+1}=\frac{y^q}p-(1-\frac 1q)y^q=0$$

Q.E.D

This is standard theorem of means. If $$(x + y) = 1$$ then $$a^{x}b^{y} \leq ax + by$$. In this equation replace $$a$$ by $$a^{p}$$ and $$b$$ by $$b^{q}$$ and $$x = 1/p$$, $$y = 1/q$$. The condition $$\dfrac{1}{p} + \dfrac{1}{q} = 1$$ is missing in the question.

The proof of general theorem of means follows by the concavity of function $$f(x) = \log x$$. Clearly $$f''(x) = -1/x^{2} < 0$$ therefore the function $$f(x) = \log x$$ is concave (meaning the chord connecting two points on graph of $$y = \log x$$ will be below the part of the graph between those two points). Algebraically this means that if $$x, y \in [0, 1]$$ with $$x + y = 1$$ then $$f(ax + by) \geq xf(a) + yf(b)$$. Thus we get $$\log(ax + by) \geq x\log a + y\log b$$ or $$a^{x}b^{y} \leq ax + by$$

• $f(ax+by)\geq xf(a)+xf(b)$ should be $f(ax+by)\geq xf(a)+{\color{red}{y}}f(b)$?
– mike
Commented Jun 7, 2023 at 9:37