Young's Inequality alternative proof

[grabthat123]

Young's Inequality can be restated as:

s^(x)t^(1-x)<=xs + (1-x)t where s,t>=0 and 0<x<1.

Basically I've been asked to prove this. I've been fiddling about with it for a couple of hours
to no avail.

I've tried to substitute t=e^u and s=e^v and then use partial differentiation w.r.t to v on st, but I'm not getting the required form. (I can't assume that exp is a convex function - otherwise it follows trivially)

Thanks in advance.

 

[susskind_leon]

I don't know how to prove it yet, but my 2 cents would be:
$$(\frac{s}{t})^x t \leq sx + (1-x)t$$
$$(\frac{s}{t})^x \leq \frac{s}{t} x + 1-x$$
$$a^x \leq a x + 1 - x$$
for $a > 0$
This way, you have one variable less to worry about

 

[susskind_leon]

Actually, think about it, consider the function
$$f(x)=ax + 1 - x - a^x$$
$$f(0)=1-1=0$$
$$f(1) = a + 1 - 1 - a = 0$$
$$f'(x_0)= a - 1 - \ln a a^{x_0} = 0$$
$$a^{x_0} = \frac{a - 1}{\ln a}$$
Now assume $a>1$. Then if you can show that the rhs is smaller than a, you have shown that $0 < x_0 < 1$. so
$$a-1 < \ln a a$$
$$1 - \frac{1}{a} < \ln a$$
I think you can find this property somewhere.
$$f''(x)=- (\ln a) ^2 a^{x}$$ which is smaller than 0 everywhere, particularly at $x_0$.
So for a > 1 you have already proven everything. For a < 1, you have to show that the rhs is larger than a. Let $a = 1/b$ with $b > 1$.
$$\frac{\frac{1}{b}-1}{\ln \frac{1}{b}} > \frac{1}{b}$$
$$\frac{\frac{1}{b}-1}{-\ln b} > \frac{1}{b}$$
$$1 - b < \ln b$$
since b > 1, this should hold trivially.
Anyhow, it all comes down to $1 - \frac{1}{a} < \ln a$. You should look it up (and make sure I haven't made any mistakes) lol

 

[susskind_leon]

$$f(a)=\ln a a +1 - a \geq 0$$
For $a = 1$ you have f(1)=0, so it's correct.
$$f'(a)= \frac{a}{a} + \ln a - 1 = \ln a$$
which is always greater than 0, so you have a monotonically growing function.