TITLE: Proof of inequality $\sum\limits_{k=0}^{n}\binom n k\frac{5^k}{5^k+1}\ge\frac{2^n\cdot 5^n}{3^n+5^n}$
QUESTION [7 upvotes]: 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 $$\binom n k=\frac{n!}{k!(n-k)!}$$

REPLY [7 votes]: 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.