grandes-ecoles 2018 Q5

grandes-ecoles · France · centrale-maths1__official Proof Direct Proof of an Inequality
Let $p$ and $q$ be two strictly positive reals such that $\frac{1}{p} + \frac{1}{q} = 1$. Show that, for all non-negative reals $a$ and $b$,
$$ab \leqslant \frac{a^{p}}{p} + \frac{b^{q}}{q}$$
You may use the concavity of the logarithm.
Let $p$ and $q$ be two strictly positive reals such that $\frac{1}{p} + \frac{1}{q} = 1$. Show that, for all non-negative reals $a$ and $b$,

$$ab \leqslant \frac{a^{p}}{p} + \frac{b^{q}}{q}$$

You may use the concavity of the logarithm.