How to prove that$D<0.5 T\;$?

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We have the formula of obligation duration:

$$D = \frac{C}{P} (1- \frac{1+Tr}{(1+r)^{T}}) \frac {1+r}{r^2} + \frac {C+N}{P} \frac{T}{(1+r)^2},$$

Here:

$r\in(0,1)$,

$T\in \mathbb N$ (number of time periods),

$N\in \mathbb N$ (obligation nominal value),

$P \in \mathbb N$ (obligation price), $C=cN, c\in(0,1).$

Check if $D<\dfrac{T}{2}.$

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Your assumption is false.

Counterexample: Define the following numbers $$r = \frac{1}{2}, $$ $$T = 1, $$ $$N = 1, $$ $$P = 1, $$ $$C = \frac{1}{2}\cdot 1 = \frac{1}{2}. $$

Then \begin{align*}D &= \frac{C}{P} \left(1- \frac{1+Tr}{(1+r)^{T}}\right) \frac {1+r}{r^2} + \frac {C+N}{P} \frac{T}{(1+r)^2}\\ &= \frac{1}{2}\left(1 - \frac{1 + 1/2}{1 + 1/2}\right)\cdot \frac{1+ 1/2}{1/4} + \frac{1/2 + 1}{1}\cdot \frac{1}{(1+ 1/2)^2}\\ &=\frac{1}{1+1/2} = \frac{2}{3}. \end{align*}

Since $$D = \frac{2}{3}> \frac{1}{2} = \frac{T}{2},$$ we concluded that $$D< \frac{T}{2}$$ is, in general, false.