Solving Inequalities: Need Help with Part (a)?

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    Inequalities
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To solve the inequality in part (a), start by expressing the difference between the two quantities involved. This foundational step will help clarify the relationship and guide further calculations. It is essential to isolate the variable to find the solution set. Seeking hints or guidance can be beneficial for understanding the process. Engaging with others can enhance problem-solving skills in inequalities.
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Homework Statement
An object, constrained to move along the x axis, travels a dis-
tance d1 with constant velocity v1 for a time t1 . It then instan-
taneously changes its velocity to a constant v2 for a time t2 ,
traveling a distance d2 . (a) Show that (v1d1+v2d2)/(d1+d2) is greater than or equal to (v1t1+v2t2)/(t1+t2).
(b) Under what conditions is this an equality?
Relevant Equations
v = v0 + at
x = x0 + v0t + (at^2/2)
I’m not sure of how begin solving for the inequality. How do I begin solving part (a)? Could I have a hint? Thank you!
 
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A good start would be to write the expression for the difference between them.
 
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Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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