# Proving inequality

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1. Feb 22, 2015

### Vriska

1. The problem statement, all variables and given/known data

this is the problem , if x and y are real positive numbers , I need to prove

$$4x^4 + 4y^3 + 5x^2 + y + 1 \ge 12xy$$

2. Relevant equations

$$x^2 + y^2 \ge 2xy$$ (Variation of AM GM Theorem)

3. The attempt at a solution
but $$x^2 + y^2 \ge 2xy$$, so $$6x^2 + 6y^2 \ge 12xy$$, from this if

$$4x^4 + 4y^3 + 5x^2 + y + 1 \ge 6x^2 + 6y^2$$ then the given inequality holds

let's try this $$44x^4 + 4y^3 \ge 6x^2 + 6y^2$$

$$4x^4 + 4y^3 / (6x^2 + 6y^2) \ge 1$$

$$4x^4 + 4y^3 /(6x^2 + 6y^2) - 1 \ge 0$$

$$4x^4 + 4y^3 - 6x^2 - 6y^2 /(6x^2 + 6y^2) \ge 0$$ , divide both sides by the numerator ( this step feels wrong)

$$1/(6x^2 + 6y^2) \ge 0$$

since this is true the inequality $$4x^4 + 4y^3 \ge 6x^2 + 6y^2$$ holds , so from this we can say that

$$4x^4 + 4y^3 + 5x^2 + y + 1 \ge 4x^4 + 4y^3$$

$$5x^2 + y +1 \ge 0$$

, this is obviously true and thus the above inequality holds

BUT , the inequality $$4x^4 + 4y^3 \ge 6x^2 + 6y^2$$ doesn't hold for x and y = 1 but it does seem to work for all values of greater than 2 . Where have I gone wrong ?
There seems to be a better way to solve the problem because solving this in such a round about way seems weird . A better solution (this proof does not work for numbers from 0 - 2 ) to this would be helpful.

2. Feb 22, 2015

### LCKurtz

Not sure if it helps you or not, but you can find the critical point of $4x^4 + 4y^3 + 5x^2 + y + 1 - 12xy$ numerically and that it occurs at

and the minimum value of the function is

3. Feb 22, 2015

### Ray Vickson

Your above point P1 is is actually a saddle point. The min in the first quadrant is at P2 = {x = .5522496847, y = .6847746720}, giving f = 0.328132751. I found these by first solving $f_y=0$ for $x$, then substituting into $f_x = 0$ to get a 6th degree polynomial in $y$ whose two positive roots are {y = .7508111365e-1} and {y = .6847746720}---Hoorray for Maple!. The Hessian of f at P1 is indefinite, while at P2 is positive definite. A 3d plot shows f > 0 in the first quadrant, and a contour plot of f near (0,0) shows clearly the saddle-point nature of P1.

4. Feb 23, 2015

### Vriska

This really doesn't help , this question was asked in a regional math olympiad to 10th graders so this information doesn't mean much to me . I need a simple proof using the AM GM theorem and also maybe a way to prove this inequality works for x and y lesser than 2 but greater than 0

5. Feb 23, 2015

### Ray Vickson

Here is an elementary argument that comes close, but does not quite do what you want. First, note that if $t_1, t_2, \ldots, t_6 > 0$ are some positive numbers and $a_1, a_2, \ldots, a_6 > 0, \; a_1 + a_2 + \cdots + a_6 = 1$ are positive "weights" summing to 1, then
$$a_1 t_1 + a_2 t_2 + \cdots + a_6 t_6 \geq t_1^{a_1} \cdot t_2^{a_2} \cdots t_6^{a_6}$$
This is a generalization of the AM/GM inequality, Apply it when the left-hand-side is written as
$$\text{LHS} = \frac{1}{6} (24 x^4) + \frac{1}{4} (16 y^3) + \frac{1}{6} (30 x^2) + \frac{1}{4} (4y) + \frac{1}{6} (6)$$
That will give you
$$\text{LHS} \geq (24 \cdot 30 \cdot 6)^{1/6} \, (16 \cdot 4)^{1/4} xy = 4320^{1/6} 64^{1/4} xy \doteq 11.41455411\, xy$$
This gives LHS >= 11.416 xy instead of LHS >= 12 xy. So, as I said: close, but not quite what you want. Again, I don't know if this helps you.

Note added in edit: perhaps it is possible to group the terms on the left in another way and with different weights, allowing use of the more general AM/GM inequality
$$\sum_{i=1}^m a_i t_i \geq \prod_{i=1}^m t_i^{a_i}$$
where $t_i, a_i > 0, i=1, \ldots m$ and $\sum a_i = 1$.

Edit note 2: we can get a constant >= 12 on the right by looking at the problem as follows: if
$$t_1 = 4x^2, \: t_2 = 4 y^3, \: t_3 = 5x^2, \: t_4 = y, \: t_5 = 1$$
we have
$$f(x,y) = t_1 + t_2 + t_3 + t_4 + t_5 \\ = a_1(t_1/a_1) + a_2 (t_2/a_2) + a_3 (t_3/a_3) + a_4 (t_4/a_4) + a_5 (t_5/a_5 )\\ \geq (t_1/a_1)^{a_1} (t_2/a_2)^{a_2} (t_3/a_3)^{a_3} (t_4/a_4) ^{a_4} (t_5/a_5)^{a_5}\\ = (4/a_1)^{a_1}(4/a_2)^{a_2} (5/a_3)^{a_3} (1/a_4)^{a_4} (1/a_5)_{a_5} x^{4a_1+ 2a_3} y^{3a_2+a_4}$$
Thus, $f(x,y) \geq c xy$, if
$$4 a_1 + 2 a_3 = 1,\\ 3 a_2 + a_4 = 1\\ a_1 + a_2 + a_3 + a_4 + a_5 = 1$$
For given $a_i$ the constant $c$ on the right is
$$c = \left( \frac{4}{a_1}\right)^{a_1} \left( \frac{4}{a_2} \right)^{a_2} \left( \frac{5}{a_3} \right)^{a_3} \left( \frac{1}{a_4} \right)^{a_4} \left( \frac{1}{a_5} \right)^{a_5}$$
There are many values of the $a_i$ that give $c \geq 12$. (In particular, if we maximize $c$ subject to the constraints on the $a_i$ we can get $c = 14.4902646505479994 > 12$. (In other words, we actually have $f(x,y) \geq 14.49 xy$.) If you do not care to use an optimization package (or are not allowed to do so) you can just try some values manually until you stumble upon a value $c \geq 12$.

It would help to use the restrictions on the $a_i$ to solve for three of them in terms of the other two; for example, we can easily solve for $a_3, a_4, a_5$ in terms of $a_1,a_2$. Then $c$ can be re-written in terms of $a_1, a_2$ only, and there are some restrictions on $a_1, a_2$ due to the requirements $a_3,a_4,a_5 \geq 0$. One can then just try some trial values of $(a_1,a_2)$ until we get a value $c \geq 12$. In principle, all you need is a good hand-held calculator.

Last edited: Feb 23, 2015
6. Feb 23, 2015

### Raghav Gupta

But @Vriska is saying that it was a regional Olympiad. In that Maple and calculators are not allowed.
It is a bit of manipulation work and not calculative one by putting values for x and y( Olympiads not want that). At the moment I am not able to solve it but will try.

7. Feb 23, 2015

### LCKurtz

You're right of course. I let Maple solve the system and it only found the saddle point. And when I did a contour plot I didn't notice the min it showed was in a different place. And I didn't plot enough contours to see the saddle point. Just shows you have to be careful with Maple. As far as that being an Olympiad problem, well, good luck with that. I have other things to spend my time on...

8. Feb 23, 2015

### Ray Vickson

Post #5 (final part) presents a method that requires no more than a hand-held "scientific" calculator. Again, good luck to the OP!

9. Feb 23, 2015

### SammyS

Staff Emeritus
Try this:

$4x^4 + 4y^3 + 5x^2 + y + 1$

$=\left(4x^4 +1 \right) +y \left(4y^2 +1\right) + 5x^2$​

Use your variation of AM/GM on each of the parenthetical quantities and proceed.

10. Feb 23, 2015

### Raghav Gupta

Brilliant.
Proceeding by your method we get this all expression is greater than 9x2 + 4y2.
Now obviously as stated by OP 6x2+ 6y2 >= 12xy
Now how we will prove that 9x2 + 4y2 is greater than 6x2+ 6y2?

11. Feb 24, 2015

### SammyS

Staff Emeritus
OP may have been wrong in that regard, but the result doesn't depend upon that. It's not necessarily true that $9x^2+4y^2\ge 6x^2+6y^2\ .$

I leave it to OP to complete the solution.

12. Feb 24, 2015

### Raghav Gupta

I have an interest in the problem,
But isn't
6x2+ 6y2 is greater than or equal to 12xy by the AM-GM inequality?

13. Feb 24, 2015

### SammyS

Staff Emeritus
Yes, but that is irrelevant to the problem at hand.

$3x^2 + 12y^2 \ge 12xy \$ also, as well as other combinations.

14. Feb 24, 2015

### Raghav Gupta

How that's true? Initially we get x2 + y2 greater than or equal to 2xy
Then how
$3x^2 + 12y^2 \ge 12xy \$ also, as well as other combinations.

15. Feb 24, 2015

### Raghav Gupta

So this is wrong in reality.
If taking x=1 and y=1
$$4x^4 + 4y^3 - 6x^2 - 6y^2$$ will be negative and when we divide a thing by negative in inequalities the inequality direction changes. Same the case for taking reciprocals.
For example,
1> 0.
When we divide by -1 on both sides,
-1<0 so the sign changes.

16. Feb 24, 2015

### Ray Vickson

The AM/GM inequality say that for $A,B > 0$ we have $A + B \geq 2 \sqrt{A B}$, so $3 x^2 + 12y^2 \geq 2 \sqrt{3 x^2 \,12 y^2} = 2 \sqrt{36} xy = 12 xy$.

17. Feb 24, 2015

### Raghav Gupta

Got it,
And from here also the answer indirectly. Thanks.
Now it's all for the OP to understand.