# Analysis problem

1. Sep 1, 2007

### ELESSAR TELKONT

My problem is the next:

Show that $$\forall x,y\geq 0$$ and $$\frac{1}{p}+\frac{1}{q}=1$$ we have that $$xy\leq\frac{x^{p}}{p}+\frac{y^{q}}{q}$$.

I use the logarithm function because it is concave and grows monotonically and then

$$\ln{xy}=\frac{\ln{x^{p}}}{p}+\frac{\ln{y^{q}}}{q}$$

the problem is I don't know to justify the next step (or is justified yet?)

$$\ln{xy}=\frac{\ln{x^{p}}}{p}+\frac{\ln{y^{q}}}{q}\leq \ln\left(\frac{x^{p}}{p}+\frac{y^{q}}{q}\right)$$.

That's all, I wait for your help.

2. Sep 1, 2007

### Hurkyl

Staff Emeritus
Derivatives are often good at proving inequalities. e.g. you can prove
f(x) < g(x)
for all positive x as follows:
. f(0) < g(0)
. f'(x) < g'(x) for all positive x

(I don't know if they will be useful here, but it's something I'd try)

3. Sep 3, 2007

### SanjeevGupta

Hi
This may be brute force but it looks like it works:
Use Arithmetic Mean>= Geometric Mean.
1/p+1/q=1 implies p+q=p.q
Also x^p/p+y^q/q=(q.x^p+p.y^q)/(p.q)
Now AM>=GM implies
(q instances of x^p plus p instances of y^q)/(p+q)>={x^(p.q).y^(q.p)}^(1/(p+q))
LHS is just (p.q).{x^p/p+y^q/q}/(p+q)=x^p/p+y^q/q
RHS is x.y

Hope this helps

Last edited by a moderator: Sep 3, 2007
4. Sep 3, 2007

### Gib Z

Hurkyl, Derivatives only help easily in inequalities of 1 variable :( Even if he expressed y as a function of x, the condition on p and q pose difficulty.

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