Question

Demonstrate that\[\frac{x^2+y^2}{4}\le e^{x+y-2}\]for all\[x\ge0,y\ge0\]

Answer 1

A push in the right direction on this would suffice.

Answer 2

...I hope.

Answer 3

isn't right side circle?

Answer 4

I suppose so. Isn't the left?

Answer 5

yes, too early

Answer 6

lol

Answer 7

right side is

\[\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\text{...}\right)\left(1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+\text{...}\right)\frac{1}{e^2}\]

Answer 8

Write a function which is the difference, f(x,y) = e^(x+y-2) - (x^2+y^2)/4

Now, what I'd do is
- Show that f(x,y) is non-negative on the axes (x,0) x ≥ 0, (0,y) y ≥ 0
- Show grad f points in the right direction; namely the x and y components are both positive, so we know f is increasing as we move away from those axes.
- Then I'd find the right theorem so that I can formalize the conclusion of that second step and hence f(x,y) cannot be zero in that domain.

If I get stuck on the last step, I might try and find a more elementary proof. I'll think about this a bit more.

Answer 9

I'm pretty sure I will get stuck there, but thanks for the start, I'll give it a try.

Answer 10

That seems equivalent to what James said. Only he used the difference of the functions.

Answer 11

\[\frac{x^2+y^2}{4}=\frac{x^2+y^2}{4}=\frac{(x+y)^2-2xy}{4}\le\frac{(x+y)^2}{4}\]

Answer 12

show \[\frac{z^2}{4}\le e^{z-2}\]...which is easy

Answer 13

Yes, that's a more elementary proof.

Answer 14

Induction?

Answer 15

use calculus

Answer 16

Write g(z) = e^(z-2) - z^2/4. Now use your elementary calculus to show that g(z) is positive for all z ≥ 0

Answer 17

[
For the record, my method is intuitively sound, but I've been scribbling on some paper figuring out to formalize it. It's a bit complicated. One needs to

a) establish the differential equations of gradient flow (d/dt)(x,y) = grad f; initial conditions f(x,0), f(0,y) [easy]

b) Show that the solutions to those equations correspond to the values of f(x,y) for every point in the domain that is attained [not so bad]

c) every point in the domain D = { (x,y) | x, y ≥ 0 } is attained by the gradient flow equations [tricky]

In short, much better to use Zarkon's approach ;-)
]

Answer 18

I figured as much, but I'm not used to doing proofs like this through calculus, so I'm rather of-balance. Easy he says...

Answer 19

off*

Answer 20

Actually, show that g(z) attains a minimum for z ≥0 and that minimum is positive.

Answer 21

the min is actually zero (at z=2)...but that is ok

Answer 22

non-negative, yes.

Answer 23

But how in the heck do I solve\[e^{z-2}-\frac{z}{2}=0\]???
hmmm...

Answer 24

It's obvious by inspection that you are right. z=2

Answer 25

there are actually two real solutions to that: http://www2.wolframalpha.com/input/?i=solve+e%5E%28x-2%29%3Dx%2F2

Answer 26

But I'm with you @TuringTest - how do we find the solution analytically here?

Answer 27

Yeah, I did that as well, but how does that help if I can't find them? I don't even know what the product log function is, is that really necessary?
Ah you'r stuck there too, at least I'm not alone :)

Answer 28

:)

Answer 29

I guess we don't have the photon drives installed in our brains - like @Zarkon and @JamesJ seem to have :-D

Answer 30

yeah, I like how they say to use elementary calc to show that g(z) has a min at z=2
I know to take the derivative, no need to explain that. I don't know how to solve it is the problem.
Here comes a tip...

Answer 31

z = 2 is a solution of g'(z) = 0 and g'(z) > 0 for all z > 2 (why?),
hence g(z) ≥0 for all z ≥ 2.

Now, you'll need to be a bit creative (Rolle's Theorem? Mean Value Theorem? Change of sign of g''?), but stick at it for a bit and show there cannot be another zero between 0 and 2.

Answer 32

That sounds promising :)

Answer 33

I went off on a tangent I proved this instead:\[\begin{align}
f&=ex-e^x\\
f'&=e-e^x\\
\text{this is zero when }x&=1\\
f''&=-e^x\\
\text{this is negative at }x&=1\\
\text{therefore we have a maximum at }x&=1\\
f_{max}&=e-e=0\\
\therefore f\le0\\
\therefore ex-e^x&\le0\\
\therefore ex&\le e^x\\
\therefore x&\le e^{x-1}\\
\therefore xy&\le e^{x-1}e^{y-1}\\
\therefore xy&\le e^{x+y-2}
\end{align}\]

Answer 34

haha, still interesting though.

Answer 35

Well I'm gonna try some of those tricks above...

Answer 36

yes - I'll give it a go as well...

Answer 37

its easy enough to show that g'(z) > 0 for all z > 2 with the second derivative, but the rest...
well let's see how creative you've been :)

Answer 38

\[\begin{align}
g&=e^{z-2}-\frac{z^2}{4}\\
g'&=e^{z-2}-\frac{z}{2}\\
g''&=e^{z-2}-\frac{1}{2}\\
\end{align}\]we know z=2 gives g'=0. at z=2 g'' is positive so this represents a minimum
\[\begin{align}
g_{min}&=e^0-\frac{4}{4}=1-1=0\\
\therefore g&\ge0\\
\therefore e^{z-2} -\frac{z^2}{4}&\ge0\\
\therefore e^{z-2} &\ge \frac{z^2}{4}
\end{align}\]
oops - should have read your reply before carrying on :-)

Answer 39

I'd still like to know how to analytically find z=2 (and z=0.406376...) are the solutions to g'=0

Answer 40

Hey that wasn't bad at all :P
I guess I need to work on these, nice job!
^^^with you on that bit too.

Answer 41

Although James made the argument that we need to prove that no other zeros exist between zero and two, did you do that?

Answer 42

no - I think we need to check the other solution as well, i.e. z=0.406376..
those are the only two that make g'=0

Answer 43

Yeah, he says to show that there isn't a zero tin [0,2) but there is...

Answer 44

in*

Answer 45

a zero of g(x) I mean

Answer 46

ah... I misunderstood.

Answer 47

at z=0.406376, g''=-0.296812 which means that represents a minimum

Answer 48

a maximum

Answer 49

sorry - I meant max

Answer 50

therefore I think we have proved the assertion?

Answer 51

But you had to use that solution that can't be found through inspection, how am I supposed to do that on the test? They don't let you bring Wolfram into the class.

Answer 52

yes - which is why I also am not 100% happy with the solution

Answer 53

g(0) = e^-2. And we know that g(2) = 0. Further we know that for x just less than 2,
g'(x) < 0.

We also know that there is a zero of g(x), call it x1 and x1 < 0. Now, if there is another zero of g(x) between x1 and 2. Show that g(x) must have both a local max and local min in that range. But if that is the case, then g''(x) must have two zeros, but it doesn't.

Answer 54

I guess this is what @JamesJ is trying to say: