# Verifying an inequality

$$\psi(t):\to[0,\infty)$$ and $$\phi(t):\to[0,\infty)$$

where;

$$\psi(t)=3/2*t^2$$ and $$\phi(t)=t$$

$$d(x,y)=|x-y|$$ where $$x,y \epsilon \mathbb{R}$$

The above are the conditions which I want to put on the given inequality

$$$$3*t^2*d(x,y)-3/2*t^2*d(x,y)]\leq(3/2*t^2*d(x,y))^\alpha*(t*d(x,y))^{1-\alpha}$$$$ where $$\alpha \epsilon(0,1)$$

I have tried the code below:

-Infinity < x < Infinity;

-Infinity < y < Infinity;

0 <= t < Infinity;

0 < alpha < 1;

d(x,y) = Abs[x-y];

3*t^2*d(x,y)-3/2*t^2*d(x,y) <= (3/2*t^2*d(x,y))^alpha*(t*d(x,y))^1-alpha

Furthermore, I have also tried it with Reduce[ ]

• What did you try with Reduce[]? – J. M. is away Mar 28 at 10:05
• No, I just simply wrote the inequality and the conditions – WKhan Mar 28 at 10:07
• Hi WKhan, welcome to Mathematica.SE! This site is about questions regarding the software Mathematica. In case you are trying to solve your inequality with Mathematica it's always good to update your question and show the code you already tried and what your specific question is. If you have a general math question math.stackexchange.com is the place for that, in that case your question should be migrated. – Thies Heidecke Mar 28 at 11:05
• The assumptions should be written as assume = Element[{x, y}, Reals] && t >= 0 && 0 < alpha < 1; The definition for d should be d[x_,y_] = Abs[x-y]. The inequality should be written ineq = 3*t^2*d[x, y] - 3/2*t^2*d[x, y] <= (3/2*t^2*d[x, y])^alpha*(t*d[x, y])^(1 - alpha). Unfortunately, Reduce[ineq && assume] does not work even with the corrections. – Bob Hanlon Mar 28 at 14:32
• Thank you for you time, I tried your given code and yeah it doesn't work with Reduce and even with Refine – WKhan Mar 28 at 14:42