I'm having some trouble deciding whether this inequality is true or not... $| \sqrt[3]{x^2} - \sqrt[3]{y^2} | \le \sqrt[3]{|x -y|^2}$ for $x, y \in \mathbb{R}.$
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2What's the relation between $u,v,x,y$? – Hagen von Eitzen Jul 28 '15 at 12:38
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Sorry by mistake, I written $u,v$. Now, the question is edited & seems correct @Hagen von Eitzen – Lakshmi Narayan Mishra Jul 28 '15 at 12:45
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Please try someone. I am also trying but I can not prove this inequality so please please help me in this question. – Lakshmi Narayan Mishra Jul 28 '15 at 13:22
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2I guess this might be useful: http://math.stackexchange.com/questions/233146/is-xt-subadditive-for-t-in-0-1 and http://math.stackexchange.com/questions/134714/does-xp-with-0p1-satisfy-the-triangular-inequality-on-mathbbr – Martin Sleziak Jul 28 '15 at 13:55
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$|x - y|^2 = (x - y)^2 $
Let $\sqrt[3]{x} = a , \sqrt[3]{y} = b$ I will study this $$| a^2 - b^2 | \leq \sqrt[3]{|a^3-b^3|^2} $$ L.H.S $$|a^2 - b^2|^3 = |a-b|^3 \cdot |a+b|^3= \color{red}{|a-b|^2}\cdot |a^2-b^2|\cdot |a^2 +2ab+b^2|$$ R.H.S $$|a^3 - b^3|^2 = \color{red}{|a-b|^2} \cdot |a^2 +ab + b^2|^2 $$
So our problem reduced into studying if $$|a^2-b^2|\cdot |a^2 +2ab+b^2| \leq |a^2 +ab + b^2|^2$$ If $a,b >0$. Then L.H.S $$\color{red}{a^4 +2a^3b} - 2ab^3 -b^4$$ R.H.S $$\color{red}{a^4 +2a^3b} + 3a^2b^2 +2ab^3 + b^4$$ which is absolutely bigger than L.H.S
Hope it will help you..
IrbidMath
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In the RHS part you have not taken $\sqrt[3]{|a^3 - b^3|^{2}}$ so please check it & try to correct it@AmerYR – Lakshmi Narayan Mishra Jul 28 '15 at 15:54
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I take the cube exponent for both sides. $|a^2 - b^2| \leq \sqrt[3]{|a^3-b^3|^2}$ I look for this $|a^2 -b^2|^3 \leq |a^3 - b^3|^2$ if i proved it then we are done. since if $x < y$ then $x^3 < y^3$. and if $r < s $ we have also $\sqrt[3]{r} < \sqrt[3]{s}$. Both the functions $x^3, \sqrt[3]{x}$ are increasing functions – IrbidMath Jul 28 '15 at 16:30
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no problem but i did not finish the question. I will try to complete it.I solved it for the case $a,b >0$ – IrbidMath Jul 28 '15 at 16:42
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The case $a,b < 0$ is the same as $a,b>0$ to see this if we let $a = -r, b = -t$ in the inequality we will get the same. So the final case is when $a>0 , b<0$ – IrbidMath Jul 28 '15 at 17:03
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Your solution is really good. Thanks for the above steps. Keep on top@AmerYR – Lakshmi Narayan Mishra Jul 29 '15 at 17:21
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It is shown in Is $x^t$ subadditive for $t \in [0,1]$? that we have the inequality $$(a+b)^t \le a^t + b^t \tag{1}$$ for all nonnegative $a,b$ and all $t \in [0,1]$.
For your inequality, let us assume without loss of generality that $x \ge y$. Applying the above inequality (1) with $a=y$, $b=x-y$, and $t=2/3$ we get $$x^{2/3} \le y^{2/3} + (x-y)^{2/3}.$$ Subtracting $y^{2/3}$ from both sides, $$x^{2/3} - y^{2/3} \le (x-y)^{2/3}.$$ Now since $x \ge y$ we have $x-y = |x-y|$. And since the function $x \mapsto x^{2/3}$ is increasing, we have $x^{2/3} \ge y^{2/3}$, so $x^{2/3} - y^{2/3} = |x^{2/3} - y^{2/3}|$. Thus we showed $$|x^{2/3} - y^{2/3}| \le |x-y|^{2/3}$$ which is the desired inequality.
Nate Eldredge
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it boils down to showing that $|1-u^{2/3}|\leq(1-u)^{2/3}$, $1<u<1$, $u=x/y$, $0<x<y<1$.This is the same as showing that $1<u^{2/3}+(1-u)^{2/3}$. Since $0<u$, $1-u<1$ we know that $u<u^{2/3}$, $1-u<(1-u)^{2/3}$. Adding up we get the affirmative result.
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After over a hundred posts, you really should be learning to use MathJax... – Micah Jul 28 '15 at 15:35