Proof of inequality $\sum\limits_{k=0}^{n}\binom n k\frac{5^k}{5^k+1}\ge\frac{2^n\cdot 5^n}{3^n+5^n}$ 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 $$\binom n k=\frac{n!}{k!(n-k)!}$$
 A: 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. 
A: 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.
