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- Feb 14, 2012

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- Thread starter anemone
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The next step is to prove by induction that $c_n \geqslant 2\sqrt{2n}$ for all $n \geqslant 2$, with strict inequality except possibly when $n=2$. The base case $n=2$ holds, because $c_2 \geqslant 4$. Suppose that the inequality holds for $n$. Since the function $f(x)$ is an increasing function for all $x\geqslant2$, it follows that $f(c_n) \geqslant f(2\sqrt{2n}).$ Therefore $$c_{n+1}^2 = (f(c_n))^2 \geqslant (f(2\sqrt{2n}))^2 = \left(2\sqrt{2n} + \dfrac4{2\sqrt{2n}}\right)^2 = 8n + 8 + \dfrac2n > 8(n+1).$$ Take square roots of each side to get $c_{n+1} > 2\sqrt{2(n+1)}$, which completes the inductive step.

Finally, take $n=50$ to see that $a_{50}+b_{50} = c_{50} > 2\sqrt{100} = 20$.