I'm going through Carothers Real Analysis to study for my qualifying exams, and came across this unjustified statement, that Youngs inequality (for $p > 1$ and q such that $\frac{1}{p} + \frac{1}{q} = 1$ we have that for $a, b \geq 0$, $ab \leq \frac{a^p}{p} + \frac{a^q}{q}$), when $p=q=2$, reduces to the arithmetic geometric mean inequality $\sqrt{ab} \leq \frac{a+b}{2}$. I tried manipulating youngs inequality but cant seem to derive the arithmetic geometric mean inequality from it. How is it done?
How does Young’s inequality reduce to the arithmetic-geometric mean inequality
real-analysis
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Best Answer
Substitute $\sqrt a$, $\sqrt b$ for $a,b$, respectively.