2
$\begingroup$

ّFind the minimum real number $\lambda$ so that the following relation holds for arbitrary real numbers $x,y$($x>y$): $$\lambda\frac{x^3-y^3}{(x-y)^3}\geq \frac{x^2-y^2}{(x-y)^2}$$

I simplified the relation into the following: $(\lambda-1)x^2+\lambda xy+(\lambda +1)y^2\geq0$ , but can't go further...

3 Answers 3

1

Recall that for a quadratic to be $\geq0$ you must have that discriminant $\Delta\leq 0$ and the leading coefficient $> 0$ in this case $\lambda>1$ and $\lambda^2y^2-4(\lambda^2-1)y^2\leq 0\iff \lambda^2-4\lambda^2+4\leq 0 $

  • 0
    You are right,but how to deal with $y$ in $\Delta$? – 2017-02-11
  • 0
    @HamidRezaEbrahimi $y^2$ is non-negative hence just divide by it,and take care of the case when $y=0$ which is trivially correct. – 2017-02-11
  • 0
    The last eq. never holds!!! unless for $\lambda=2$ – 2017-02-11
  • 1
    @HamidRezaEbrahimi Not true,it holds exactly when $\lambda \geq\frac{2}{\sqrt{3}}$ and $\lambda\leq -\frac{2}{\sqrt{3}}$,though if you want I'll show you to steps to solve the last inequality as well. – 2017-02-11
  • 0
    Sorry,I thought the second $\lambda$ has power 1 – 2017-02-11
  • 0
    @Test123 How about if $z$ is negative,try for example $z=-2-\sqrt{3}$. – 2017-02-11
  • 0
    I donwvoted an accepted answer for this reason : $\lambda>1$ is only a sufficient condition for the inequality to hold, while question from the OP was about the minimum acceptable value. – 2017-02-12
  • 0
    @zwim Did you read my post thoroughly?Because the $\lambda>1$ is a necessary condition and not sufficient; it's sufficient that $\lambda>1$ *AND* $-3\lambda^2+4\leq 0$,combining this two inequalities on can find the minimal $\lambda$. – 2017-02-12
  • 0
    I did, the combining condition is just $\lambda>1$. But if you read the post by victoria, you'll see that $\lambda\ge\frac{2}{\sqrt 3}$ holds as a sufficient condition and is reached for $y=(\sqrt 3-2)x, x\ge 0$. If you had something similar in mind, it is not in your post. – 2017-02-12
  • 0
    @zwim My post is $2$ lines long,and if you can't follow that much,or even read the comments of my post,then feel free to down vote,though my answer is 100% accurate. – 2017-02-12
1

Your work is interesting and I have a possible solution using that method,

First I'd like to check out this approach. I would always factor first. I don't know if you did that so I'm working it out. We are given that $x > y$ so $(x-y) >0 $ and all divisions by $(x-y)$ are defined. Multiplication by $(x-y)$ doesn't change the inequality.

$$ \lambda \frac{x^3 - y^3}{(x-y)^3} \geq \frac{x^2 - y^2}{(x-y)^2}$$

$$ \lambda \frac{(x - y)(x^2 + xy + y^2}{(x-y)^3} \geq \frac{(x-y)(x+y)}{(x-y)^2}$$

$$ \lambda (x^2 + xy + y^2) \geq (x-y)(x+y)$$

$$ (\lambda - 1) x^2 + \lambda xy + (\lambda + 1)y^2 \geq 0$$

Right, we agree here. :-)

So here's a technique used in finding limits of functions of two variables:

I got down to a solution with this, below; needs completed calculation and check as this is first draft rough work.

Let $y = kx$. Since $x > y$ given in hypothesis, if $x$ is positive k < 1 (negative is fine) and if $x$ is negative, $k < -1$. This covers all cases except $x = 0$. If $x = 0$, we calculate separately.

If $x = 0 $ your equation simplifies to $$(\lambda + 1)y^2 \geq 0$$ $y < 0 $ given $y < x$ but $$y^2 > 0$$. Therefore $(\lambda + 1) \geq 0$ and $\lambda \geq -1$

For y = kx,

$$ (\lambda - 1) x^2 + \lambda xy + (\lambda + 1)y^2 \geq 0$$ $$ (\lambda - 1) x^2 + \lambda x(kx) + (\lambda + 1)(kx)^2 \geq 0$$

$$ \lambda x^2 - x^2 + k \lambda x^2 + \lambda k^2 x^2 + k^2 x^2 \geq 0$$

Separating out the terms with $\lambda$

$$ \lambda x^2 + k \lambda x^2 + \lambda k^2 x^2 \geq x^2 - k^2 x^2$$

$$\lambda x^2(1 + k + k^2) \geq x^2(1 - k^2)$$

We dealt with $x = 0$ above. For $x \neq 0$ we can divide

$$\lambda \geq \frac{(1 - k^2)}{(1 + k + k^2)}$$

Minimize with respect to k subject to the restrictions above [if x is positive k < 1 (negative is fine) and if x is negative, $k<−1$]

$$\frac{(-2k)(1 + k + k^2) -(1-k^2)(1 + 2k)}{(1 + k + k^2)^2} = 0$$

Denominator is never zero

$$(-2k)(1 + k + k^2) -(1-k^2)(1 + 2k) = 0$$

$$ -2k -2k^2 - 2k^3 -1 - 2k + k^2 + 2k^3 = 0$$

$$ -k^2 -4k -1 = 0$$

$$ k = \frac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3}$$

$$\lambda \geq \frac{(1 - k^2)}{(1 + k + k^2)}$$

Work it out but remember the restrictions on k above.

Again rough work, not proofread, correct as needed. thanks.

Note that if $x = 0$ and $y < 0$ (allowable by definition above) then using either the simplified form or the original form we get $\lambda = -1$. This disproves some of the answers given.

My calculation above for $k$ when $y = kx , y < x$, may allow even lower value of $\lambda$. I leave it to you to take further for now.

  • 0
    Almost the best answer but you didn't conclude so well. we have $x\ge 0$ then $k\le 1$ and $x\le 0$ then $k\ge 1$. On the other hand $k1=-2-\sqrt 3$ is a minimum and $k2=-2+\sqrt 3$ is a maximum which value is $m=\frac{2}{\sqrt 3}$. So combining all restrictions ($\lambda\ge-1$ for $x=0$, $\lambda\ge m$ for $x\ge 0$ and for $x\le 0$ the condition $k\ge 1$ doesn't restrict $\lambda$ more than what we have already) then $\lambda\ge m$ is the final answer considering we ask this to work for any $x>y$. – 2017-02-12
  • 0
    I always prefer shorter solutions,also it seems your solution doesn't lead to $\frac{2\sqrt{3}}{3}$ which is valid – 2017-02-12
  • 0
    Well, Hamid, I found and proved a solution $\lambda = -1$ when x = 0 and $y < 0 \ $ that is lower than other "minimum" solutions proposed , which disproves those claims. I always remind students: There is no value in a fast mistake. Would you prefer short but wrong, or long but correct? Long can always be shortened a bit once you know where you're going.\\ $2\sqrt3 / 3$ -- since that value is positive, $\lambda = -1$ is still the lowest demonstrated value. – 2017-02-13
0

For $\lambda$, any value above $0$ is possible.

$\lambda\dfrac{x^3-y^3}{(x-y)^3}\geq \dfrac{x^2-y^2}{(x-y)^2}$

$\lambda\dfrac{x^3-y^3}{(x-y)}\geq (x^2-y^2)$

$\lambda (x^2+xy+y^2) \geq x^2-y^2$

$\lambda \geq \dfrac{x^2-y^2}{x^2+xy+y^2} \cdots(1)$

$(x-y)^2 \gt 0 \implies x^2+y^2 \gt 4xy$

Also $(x^2-y^2) \gt 0$

Since numerator and denominator of equation $(1)$ is postive, $\lambda$ is positive.

As $x$ becomes close to $y$, $\lambda$ tends to $0$

  • 0
    The RHS of (1) tends to zero. That does not tell anything about $\lambda$. – 2017-02-11
  • 0
    @MartinR, can't we assume that any value $\lambda \gt 0$ is possible, for example, for $y=1$ we get $\lambda \geq \dfrac{x^2-1}{x^2+x+1} $ and as $x$ becomes closer to $1$, $\lambda$ takes lesser value, but grater than zero? – 2017-02-11