- #1
ELESSAR TELKONT
- 44
- 0
My problem is the next:
Show that [tex]\forall x,y\geq 0[/tex] and [tex]\frac{1}{p}+\frac{1}{q}=1[/tex] we have that [tex]xy\leq\frac{x^{p}}{p}+\frac{y^{q}}{q}[/tex].
I use the logarithm function because it is concave and grows monotonically and then
[tex]\ln{xy}=\frac{\ln{x^{p}}}{p}+\frac{\ln{y^{q}}}{q}[/tex]
the problem is I don't know to justify the next step (or is justified yet?)
[tex]\ln{xy}=\frac{\ln{x^{p}}}{p}+\frac{\ln{y^{q}}}{q}\leq \ln\left(\frac{x^{p}}{p}+\frac{y^{q}}{q}\right)[/tex].
That's all, I wait for your help.
Show that [tex]\forall x,y\geq 0[/tex] and [tex]\frac{1}{p}+\frac{1}{q}=1[/tex] we have that [tex]xy\leq\frac{x^{p}}{p}+\frac{y^{q}}{q}[/tex].
I use the logarithm function because it is concave and grows monotonically and then
[tex]\ln{xy}=\frac{\ln{x^{p}}}{p}+\frac{\ln{y^{q}}}{q}[/tex]
the problem is I don't know to justify the next step (or is justified yet?)
[tex]\ln{xy}=\frac{\ln{x^{p}}}{p}+\frac{\ln{y^{q}}}{q}\leq \ln\left(\frac{x^{p}}{p}+\frac{y^{q}}{q}\right)[/tex].
That's all, I wait for your help.