Reverse AM-GM

Algebra Level 3

Find the smallest positive real number $c$ such that for all non-negative real numbers $x,y$ , we have $\sqrt{xy}+c|x-y|\ge\frac{x+y}2.$

Bonus: Generalize this for $n$ non-negative real numbers.

The answer is 0.5.

1 solution

Simon Kaib
Jul 22, 2019

At $x=y$ we have equality independent of $c$ .

Without loss of generality, let $x > y$ . Then we have: $\sqrt{xy}+c(x-y)\geq\frac{x+y}{2}$ $2c(x-y)\geq x+y-2\sqrt{xy}$ $2c(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})\geq (\sqrt{x}-\sqrt{y})^{2}$ $2c\geq \frac{(\sqrt{x}-\sqrt{y})}{(\sqrt{x}+\sqrt{y})}$ The RHS can be at most $1$ , because the denominator is $\geq$ than the numerator with equality at $y=0$ . And thus the minimum of $c$ is at $2c=1$ and we have a solution of $\boxed{0.5}$

Actually you have flipped the inequality sign

Meet Patel - 1 year, 10 months ago

Fixed, thanks

Simon Kaib - 1 year, 10 months ago

So this is assuming that $x\ne y$ ?

Ruilin Wang - 1 year, 10 months ago

If $x=y, \ |x-y|=0, \ \text{and} \ \sqrt{x^2}=x=\frac{x+x}{2}$

Théo Leblanc - 1 year, 10 months ago

Yay, thanks!!!

Ruilin Wang - 1 year, 9 months ago

This does not prove that $c=1/2$ is the minimal value for $c$ . Indeed you need to show the supremum of your lower bound $\frac{\sqrt{x}-\sqrt{y}}{ \sqrt{x}+\sqrt{y}}$ is 1. It is the case by fixing $y$ and letting $x\to +\infty$ .

Théo Leblanc - 1 year, 9 months ago

Is this enough? If not, could you please explain your point again? In case I misunderstood.

Simon Kaib - 1 year, 9 months ago

I will give you n example:

Suppose that (it is not true but it is just an example) $\frac{\sqrt{x}-\sqrt{y}}{ \sqrt{x}+\sqrt{y}}$ is never bigger than $0.6$ . Therefore, $c=1/2$ makes the inequality true but it is not optimal, $c=0.3$ works too. That is why you need to show that if $c$ is a little bit smaller than $1/2$ , the inequality is not true for all $(x,y)$ . To do so, proving that the supremum of $\frac{\sqrt{x}-\sqrt{y}}{ \sqrt{x}+\sqrt{y}}$ is 1 (ie that $\frac{\sqrt{x}-\sqrt{y}}{ \sqrt{x}+\sqrt{y}}$ can be arbitrarily close to $1$ even if it is always less than $1$ ) is enough. Indeed if you take $c<1/2$ , make $1\geq \frac{\sqrt{x}-\sqrt{y}}{ \sqrt{x}+\sqrt{y}}>\frac{1+c}{2}$ and the inequality if false.

Théo Leblanc - 1 year, 9 months ago

@Théo Leblanc – Indeed. But I did show that the supremum of the expression is $1$ by showing that it can not be more than $1$ and showing that it can be $1$ . Sorry if I got this wrong.

Simon Kaib - 1 year, 9 months ago

@Simon Kaib – Ha yes... Oups I think I miss yours words "with equality at y=0"... Your solution is perfect, sorry...

I although thought that $(x,y)>0$ that's why I talked about supremum and not maximum, my bad.

Théo Leblanc - 1 year, 9 months ago

Reverse AM-GM

The answer is 0.5.

1 solution

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