For $d \geq 2$, let
$$
f_{d}(t)=\frac{d-1}{d} \cdot t^{(d-1) / d}+\frac{1}{d} \cdot t^{-1 / d}, \quad t \geq 1
$$
Show that $f_{d}^{\prime}(t) \leq(d-1)^{2} / d^{2}$ for $t \geq 1$. Conclude that
$$
f_{d}(t)-1 \leq\left(\frac{d-1}{d}\right)^{2}(t-1), \quad t \geq 1 .
$$
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Note first $f_{d}(1)=1$ and
$$
f_{d}^{\prime}(t)=\left(\frac{d-1}{d}\right)^{2} t^{-1 / d}-\frac{1}{d^{2}} t^{-(d+1) / d} \leq\left(\frac{d-1}{d}\right)^{2}, \quad t \geq 1 .
$$
By the mean value theorem,
$$
f_{d}(t)-1=f_{d}(t)-f_{d}(1) \leq\left(\frac{d-1}{d}\right)^{2}(t-1), \quad t \geq 1
$$
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