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Show that $$\sum_{k=0}^{n}\binom n k\frac{5^k}{5^k+1}\ge\frac{2^n\cdot 5^n}{3^n+5^n}$$

where $$\binom n k=\frac{n!}{k!(n-k)!}$$

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2 Answers 2

up vote 6 down vote accepted

Note that the LHS is $$ (\ast)=\sum_{k=0}^n{n\choose k}\frac1{1+5^{-k}}=2^n\cdot\mathbb E\left(\frac1{1+5^{-S_n}}\right), $$ where the random variable $S_n$ is binomial $\left(n,\frac12\right)$. Since the function $x\mapsto\frac1{1+x}$ is convex on $x\geqslant0$, Jensen's inequality yields $$ (\ast)\geqslant\frac{2^n}{1+\mathbb E\left(5^{-S_n}\right)}. $$ Now, $S_n$ is distributed as $X_1+\cdots+X_n$ for some i.i.d. Bernoulli random variables $(X_k)_{1\leqslant k\leqslant n}$ such that $\mathbb P(X_k=0)=\mathbb P(X_k=1)=\frac12$. One gets $\mathbb E\left(5^{-X_k}\right)=\frac12\left(1+\frac15\right)=\frac35$ and, by independence, $\mathbb E\left(5^{-S_n}\right)=\left(\frac35\right)^n$. This proves the claim.

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Nice.:-D During my student years I studied much more analysis than the theory of probability. :-) –  Alex Ravsky Jun 26 '13 at 20:24
    
It's nice, Thank you +1 –  math110 Jun 27 '13 at 3:15

It seems that we can prove the inequality as follows: $$S=\sum_{k=0}^{n}\binom n k\dfrac{5^k}{5^k+1}=$$ $$\frac 12+\sum_{k=1}^{n}\binom n k\dfrac{1}{1+5^{-k}}=$$ $$\frac 12+\sum_{k=1}^{n}\binom n k\sum_{i=0}^\infty (-1)^i(5^{-k})^i=$$ $$\frac 12+\sum_{i=0}^\infty \sum_{k=1}^{n}\binom n k (-1)^i5^{-ki}=$$ $$\frac 12+\sum_{i=0}^\infty \sum_{k=1}^{n}\binom n k (-1)^i5^{-ik}=$$ $$\frac 12+\sum_{i=0}^\infty (-1)^i((1+5^{-i})^n-1).$$

Since the last series is alternating, its sum is greater than the sum of its four first members. Therefore $$S\ge \frac 12+(2^n-1)-\left(\left(\frac 65\right)^n-1\right)+\left(\left(\frac {26}{25}\right)^n-1\right)-\left(\left(\frac {126}{125}\right)^n-1\right)=$$ $$2^n-\left(\frac 65\right)^n+\left(\frac {26}{25}\right)^n-\left(\frac {126}{125}\right)^n+\frac 12.$$

Check when $$2^n-\left(\frac 65\right)^n+\left(\frac {26}{25}\right)^n-\left(\frac {126}{125}\right)^n\ge \frac{2^n\cdot 5^n}{3^n+5^n}.$$

After reducing both sides to the common denominator and simplifying we obtain the inequality $$650^n+390^n\ge 630^n+450^n+378^n.$$ But $$650^6>630^6+450^6+378^6,$$ thus the inequality holds for each $n\ge 6$.

It rests to check the initial inequality for $n\le 5$, which can be done straightforward.

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The last part is even easier to show: on LHS you have $(5 \cdot 26)^n=130^n$, on RHS $90^n-78^n$, which is clearly less than LHS –  Alex Jun 26 '13 at 16:20
    
It's nice! Thank you +1 –  math110 Jun 26 '13 at 16:25
1  
The first term in line 2 should be $\frac 1 2$ instead of 1, is not so? –  user64494 Jun 26 '13 at 17:18
    
@user64494 Yes. Thanks. I'll correct this. –  Alex Ravsky Jun 26 '13 at 18:01
    
When the last term one keeps in an alternating series is positive, the sum of the series is smaller than the truncated sum, not larger. –  Did Jun 26 '13 at 19:04

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