Lower bound for a conditional expectation [closed]

exponential-distribution conditional-expectation

From memorylessness of the exponential distribution the left-hand side is equal to $E[(2/3)^{X+6}]$ so it remains to show $$E[(2/3)^X] \ge (2/3)^4.$$

Direct approach. With $\lambda=1/4$, $$E[(2/3)^X] = \int_0^\infty (2/3)^x \lambda e^{-\lambda x} \, dx = \lambda \int_0^\infty e^{-(\lambda + \log(3/2))x} \, dx = \frac{\lambda}{\lambda + \log (3/2)} = \frac{1}{1+4\log(3/2)}$$ and you can check numerically that this is larger than $(2/3)^4$.

I may be overlooking a simpler approach for proving the inequality though.

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