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Confusion about inequality

ineq confused.png
I'm confused on why the inequality is strict. As tt+1=1−1t+1\frac{t}{t+1}=1-\frac{1}{t+1}, the expression takes a maximum value of 12\frac{1}{2} for t on the interval [ 0,1] [\ 0,1]\ . tn(t+1)n+1=tt+1⋅tn−1(t+1)n≤12⋅tn−1(t+1)n\frac{t^{n}}{(t+1)^{n+1}}=\frac{t}{t+1}\cdot \frac{t^{n-1}}{(t+1)^{n}}\leq \frac{1}{2} \cdot \frac{t^{n-1}}{(t+1)^{n}}, but they've given it as a strict inequality. Why is this?
(edited 2 years ago)
Reply 1
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mqb2766
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Original post by username53983805
ineq confused.png
I'm confused on why the inequality is strict. As tt+1=1−1t+1\frac{t}{t+1}=1-\frac{1}{t+1}, the expression takes a maximum value of 12\frac{1}{2} for t on the interval [ 0,1] [\ 0,1]\ . tn(t+1)n+1=tt+1⋅tn−1(t+1)n≤12⋅tn−1(t+1)n\frac{t^{n}}{(t+1)^{n+1}}=\frac{t}{t+1}\cdot \frac{t^{n-1}}{(t+1)^{n}}\leq \frac{1}{2} \cdot \frac{t^{n-1}}{(t+1)^{n}}, but they've given it as a strict inequality. Why is this?

The basic inequality is = only when t=1. So the integrand contains areas (0<=t<1) which are strictly less than this value.
(edited 2 years ago)
Original post by mqb2766
The basic inequality is = only when t=1. So the integrand contains areas (0<=t<1) which are strictly less than this value.

ah i see, thank you
Reply 3
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mqb2766
21
Original post by username53983805
ah i see, thank you

Sometimes is easier to just think of an integral as an average or expected value of a function (multiplied by the width of the integration interval which is 1 here). The average value here must be a strict inequality as the function only achieves its max value at one point.
(edited 2 years ago)

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