Prove that for $x,y \geq 0$ and $p>1$ , $xy \leq \frac {x^p} {p} +\frac {y^q} {q}$ whenever $ \frac {1} {p} + \frac {1} {q} = 1$.
My attempt $:$
Let us consider a function $f : (0,\infty) \longrightarrow \mathbb {R}$ defined by $f(x) = \frac {x^p} {p}$ , $x \in (0,\infty)$ where $0< p < \infty$. Then we have $f'(x)=x^{p-1}>0$ , $x \in (0,\infty)$.This shows that the function $f$ is strictly increasing in its entire domain.Now it can be easily observed that if $x=0$ or $y=0$ then the above inequality holds trivially.So WLOG let us assume that $x>0$ and $y>0$.If $x,y \geq 1$ then from the fact that $f$ is strictly increasing we must have $\frac {x^p} {p} +\frac {y^q} {q} \geq \frac {x^{\frac {p} {2}}} {p} + \frac {y^{\frac {q} {2}}} {q}$.
Now how can I proceed?But I find trouble to prove the remaining part of the inequality.Please help me.
EDIT $:$
$\frac {x^p} {p} +\frac {y^q} {q} = \frac {qx^{p} + py^{q}} {pq} = \frac {p+q} {pq} .\frac {qx^{p} + py^{q}} {p+q}=\frac {qx^{p} + py^{q}} {p+q} \geq (xy)^{\frac {1} {\frac {1}{p} + \frac {1} {q}}}=xy$, by the weighted AM-GM inequality and by using the fact that $\frac {1} {p} + \frac {1} {q} =1$.
Is my edit correct at all?Please verify it.
Thank you in advance.
