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Inequality - solve in at least two ways

lfdahl

Well-known member
Nov 26, 2013
719
Find in at least two different ways the smallest $\alpha$, such that

\[\sqrt[3]{x}+\sqrt[3]{y} \leq \alpha \sqrt[3]{x+y}\]

- for all $x,y \in \mathbb{R}_+$
 

Olinguito

Well-known member
Apr 22, 2018
251
I’ve found one way; I’ll think of another.

We note that $\sqrt[3]8+\sqrt[3]8=4=\sqrt[3]{64}>\sqrt[3]{8+8}$ and so $\alpha\ne1$. Indeed it is clear that we need $\alpha>1$.

Write $X=\sqrt[3]x,\,Y=\sqrt[3]y$. Then we want
$$(X+Y)^3\le\alpha^3(X^3+Y^3)=\alpha^3(X+Y)(X^2-XY+Y^2)$$

$\implies\ (X+Y)^2\le\alpha^3(X^2-XY+Y^2)$

$\implies\ 0\le(\alpha^3-1)X^2+(\alpha^3-2)XY+(\alpha^3-1)Y^2$

$\implies\ 0\le X^2+\dfrac{\alpha^3-2}{\alpha^3-1}XY+Y^2\le\left(\dfrac{2-\alpha^3}{1-\alpha^3}+\dfrac12\right)(X^2+Y^2)$ by AM–GM.

Since $X^2+Y^2\ge0$, we want $\dfrac{\alpha^3-2}{\alpha^3-1}+\dfrac12\ge0$. Hence
$$2(\alpha^3-2)+\alpha^3-1\ge0$$

$\implies\ \alpha\ \ge\ \sqrt[3]{\dfrac53}$.
 

lfdahl

Well-known member
Nov 26, 2013
719
I’ve found one way; I’ll think of another.

We note that $\sqrt[3]8+\sqrt[3]8=4=\sqrt[3]{64}>\sqrt[3]{8+8}$ and so $\alpha\ne1$. Indeed it is clear that we need $\alpha>1$.

Write $X=\sqrt[3]x,\,Y=\sqrt[3]y$. Then we want
$$(X+Y)^3\le\alpha^3(X^3+Y^3)=\alpha^3(X+Y)(X^2-XY+Y^2)$$

$\implies\ (X+Y)^2\le\alpha^3(X^2-XY+Y^2)$

$\implies\ 0\le(\alpha^3-1)X^2+(\alpha^3-2)XY+(\alpha^3-1)Y^2$

$\implies\ 0\le X^2+\dfrac{\alpha^3-2}{\alpha^3-1}XY+Y^2\le\left(\dfrac{2-\alpha^3}{1-\alpha^3}+\dfrac12\right)(X^2+Y^2)$ by AM–GM.

Since $X^2+Y^2\ge0$, we want $\dfrac{\alpha^3-2}{\alpha^3-1}+\dfrac12\ge0$. Hence
$$2(\alpha^3-2)+\alpha^3-1\ge0$$

$\implies\ \alpha\ \ge\ \sqrt[3]{\dfrac53}$.


Hi, Olinguito !

I think, there is a small error in the third line, - should be:


$$ 0 \leq (\alpha^3-1)X^2-(\alpha^3+2)XY + (\alpha^3-1)Y^2 $$
 

Olinguito

Well-known member
Apr 22, 2018
251
Thanks, Ifdahl .

Yes, there was an error – so I haven’t found anything after all. (Sadface) Oh well, will keep trying.
 

lfdahl

Well-known member
Nov 26, 2013
719
Hint:


Jensens inequality or power mean inequality or rearrangement inequality
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,714
With that hint...

According to the power inequality we have:
$$\frac{X+Y}{2} \le \sqrt[3]{\frac{X^3+Y^3}{2}}$$
with equality iff $X=Y$.
Let $x=X^3$ and $y=Y^3$.
Then:
$$\frac{\sqrt[3]x+\sqrt[3]y}{2} \le \sqrt[3]{\frac{x+y}{2}} \quad\Rightarrow\quad
\sqrt[3]x+\sqrt[3]y \le \sqrt[3]4\sqrt[3]{x+y}$$
Thus:
$$\alpha=\sqrt[3]4$$
 

Olinguito

Well-known member
Apr 22, 2018
251
Jensen’s inequality! Why didn’t I think of that? Stupid me. (Fubar)

The function $f:\mathbb R^+\to\mathbb R^+; t\mapsto\sqrt[3]t$ is concave, so by Jensen’s inequality for concave functions:

$$\frac{\sqrt[3]x+\sqrt[3]y}2\ \le\ \sqrt[3]{\frac{x+y}2}$$

$\implies\ \sqrt[3]x+\sqrt[3]y\ \le\ 2^{\frac23}\sqrt[3]{x+y}.$

Equality is attained when $x=y$; hence $\alpha=2^{\frac23}$.
 

lfdahl

Well-known member
Nov 26, 2013
719
With that hint...

According to the power inequality we have:
$$\frac{X+Y}{2} \le \sqrt[3]{\frac{X^3+Y^3}{2}}$$
with equality iff $X=Y$.
Let $x=X^3$ and $y=Y^3$.
Then:
$$\frac{\sqrt[3]x+\sqrt[3]y}{2} \le \sqrt[3]{\frac{x+y}{2}} \quad\Rightarrow\quad
\sqrt[3]x+\sqrt[3]y \le \sqrt[3]4\sqrt[3]{x+y}$$
Thus:
$$\alpha=\sqrt[3]4$$
Thankyou, Klaas van Aarsen for your participation and the neat solution, which of course is correct!(Yes)
 

lfdahl

Well-known member
Nov 26, 2013
719
Jensen’s inequality! Why didn’t I think of that? Stupid me. (Fubar)

The function $f:\mathbb R^+\to\mathbb R^+; t\mapsto\sqrt[3]t$ is concave, so by Jensen’s inequality for concave functions:

$$\frac{\sqrt[3]x+\sqrt[3]y}2\ \le\ \sqrt[3]{\frac{x+y}2}$$

$\implies\ \sqrt[3]x+\sqrt[3]y\ \le\ 2^{\frac23}\sqrt[3]{x+y}.$

Equality is attained when $x=y$; hence $\alpha=2^{\frac23}$.
Thankyou, Olinguito ! for your participation. Your solution is of course also correct. I also want to thank you for your nice attempt earlier. I have tried to elaborate on your solution path:


I want to thank Olinguito , for the following solution.

Let $a = \sqrt[3]{x}, \: \: b = \sqrt[3]{y}$:


\[ \left ( a+b\right )^3\leq \alpha ^3\left ( a^3+b^3 \right )=\alpha ^3(a+b)\left ( a^2-ab+b^2 \right ) \\\\ \left ( a+b\right )^2\leq \alpha ^3\left ( a^2-ab+b^2 \right )\\\\ (\alpha ^3-1)a^2+(\alpha ^3-1)b^2-(\alpha ^3+2)ab \geq 0 \\\\ a^2+b^2-\frac{\alpha ^3+2}{\alpha ^3-1}ab \geq 0\]
Hence, we see, that the fraction: $\frac{\alpha ^3+2}{\alpha ^3-1} \leq 2$

or $\alpha^3 \geq 4$, or $\alpha \geq \sqrt[3]{4}$.