<s>[INST] How prove this inequality $\sum_{k=1}^{n}\frac{2k-1}{k\binom{n}{k}}\ge \frac{n}{2^{n-1}}$ 
let $1\le k\le n,k,n\in N^{+}$, show that
  $$\sum_{k=1}^{n}\dfrac{2k-1}{k\binom{n}{k}}\ge \dfrac{n}{2^{n-1}}$$

I know this $$\sum_{k=1}^{n}(2k-1)=n^2$$
and $$\sum_{k=1}^{n}k\binom{n}{k}=n\cdot 2^{n-1}$$
I want Use Cauchy-Schwarz inequality .
$$\left(\sum_{k=1}^{n}\dfrac{2k-1}{k\binom{n}{k}}\right)\left(\sum_{k=1}^{n}k\binom{n}{k}\right)\ge (\sum_{k=1}^{n}\sqrt{2k-1})^2$$
then
$$\sum_{k=1}^{n}\dfrac{2k-1}{k\binom{n}{k}}\ge\dfrac{(\sum_{k=1}^{n}\sqrt{2k-1})^2}{n\cdot 2^{n-1}}$$
Now we must prove
$$\sum_{k=1}^{n}\sqrt{2k-1}\ge n?$$
maybe can use integral inequality to prove it.
I can't prove this.Thank you
 [/INST] First prove that: for $x≥0$ and $y≥0$, $(x+y)^{\frac{1}{2}}≤x^{\frac{1}{2}}+y^{\frac{1}{2}}$. Define $f(x)=(x+1)^{\frac{1}{2}}-x^{\frac{1}{2}}-1$. Note that $f'(x)<0$. Then for $x >0$, we have $f(x)<0$. Therefore $(x+1)^{\frac{1}{2}}<x^{\frac{1}{2}}+1$. Replace $x$ by $x/y$ we have,  $(x+y)^{\frac{1}{2}}<x^{\frac{1}{2}}+y^{\frac{1}{2}}$. As the result holds for $x=y=0$, we have $(x+y)^{\frac{1}{2}}≤x^{\frac{1}{2}}+y^{\frac{1}{2}}$, for $x≥0$ and $y≥0$. For induction,
$$(x_{1}+...+x_{n})^{\frac{1}{2}}≤x^{\frac{1}{2}}_{1}+...+x^{\frac{1}{2}}_{n}.$$
So, for $x_{k}={2k-1}$ we have, $\sum_{k=1}^{n}\sqrt{2k-1}≥\sqrt{\sum_{k=1}^{n}{2k-1}}=\sqrt{n^{2}}=n.$
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