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

• Vriska
In summary, the problem is to prove that for real positive numbers x and y, the inequality 4x^4 + 4y^3 + 5x^2 + y + 1 ≥ 12xy holds. Using the variation of AM GM Theorem, it can be shown that 4x^4 + 4y^3 + 5x^2 + y + 1 ≥ 6x^2 + 6y^2. By manipulating this inequality, it can be simplified to 4x^4 + 4y^3 ≥ 6x^2 + 6y^2. However, this inequality is not always true for values of x and y less than 2.
Vriska

Homework Statement

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

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

Homework Equations

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

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.

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
1.036443430 in the first quadrant.

LCKurtz said:
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
View attachment 79478
and the minimum value of the function is
1.036443430 in the first quadrant.

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.

LCKurtz said:
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
View attachment 79478
and the minimum value of the function is
1.036443430 in the first quadrant.

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

Vriska said:
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

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:
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.

Ray Vickson said:
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.

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...

LCKurtz said:
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...

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

Vriska said:

Homework Statement

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

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

Homework Equations

[/B]
$$x^2 + y^2 \ge 2xy$$ (Variation of AM GM Theorem)
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.

Vriska and Raghav Gupta
SammyS said:
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.
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?

Raghav Gupta said:
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?
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.

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

Raghav Gupta said:
I have an interest in the problem,
But isn't
6x2+ 6y2 is greater than or equal to 12xy by the AM-GM inequality?
Yes, but that is irrelevant to the problem at hand.

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

SammyS said:
## 3x^2 + 12y^2 \ge 12xy \ ## also, as well as other combinations.
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.

Vriska said:

The Attempt at a Solution

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

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

Raghav Gupta said:
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.

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##.

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

1. How do you prove the inequality?

The most common way to prove an inequality is by using mathematical techniques such as algebra, calculus, or graphing. In this case, we can use algebraic manipulation to rearrange the equation and show that the left side is always greater than or equal to the right side.

2. What is the significance of the inequality?

Inequalities are important in mathematics because they allow us to compare two quantities and determine which one is larger or smaller. In this case, the inequality can help us determine the values of x and y that satisfy the equation and the values that do not.

3. How do you know when the inequality is true?

To determine if the inequality is true, we can substitute different values for x and y and see if the left side is always greater than or equal to the right side. We can also use mathematical techniques to solve for the minimum or maximum values of x and y that satisfy the inequality.

4. Can you graph the inequality?

Yes, we can graph the inequality by plotting points that satisfy the equation and those that do not. The points that lie on or above the graph represent the solutions to the inequality, while the points below the graph do not.

5. Can you prove the inequality for all values of x and y?

Yes, we can prove the inequality for all values of x and y by showing that the left side is always greater than or equal to the right side, regardless of the specific values chosen. This is known as a universal proof and is considered the strongest type of proof.

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