How prove this inequality $\frac{1}{n}\sum_{k=0}^{n-1}\binom{k}{a}\binom{k}{b}\le\frac{1}{a+b+1}\binom{n}{a}\binom{n}{b}$ 
let $a,b,n$ be positive integer numbers,and such $a,b\le n$, show that
  $$\dfrac{1}{n}\sum_{k=0}^{n-1}\binom{k}{a}\binom{k}{b}\le\dfrac{1}{a+b+1}\binom{n}{a}\binom{n}{b}$$

this inequality maybe can use Integral inequality to solve it.because 
$$\int_{0}^{1}x^{a+b}dx=\dfrac{1}{a+b+1},\int_{0}^{1}x^{n-1}dx=\dfrac{1}{n}$$
Thank you for you help.

Since the inequality is symmetric in $a$ and $b$ we may (and will), suppose $a\leq b\leq n$.
Moreover, the inequality is trivially satisfied if $n=b$. So let us suppose that the inequality is proved for some $n\geq b$. It follows that
$$\eqalign{
\sum_{k=0}^n\binom{k}{a}\binom{k}{b}&=\sum_{k=0}^{n-1}\binom{k}{a}\binom{k}{b}+
\binom{n}{a}\binom{n}{b}
\cr
&\leq \frac{n}{a+b+1}\binom{n}{a}\binom{n}{b}+\binom{n}{a}\binom{n}{b}\cr
&\leq \frac{a+b+n+1}{a+b+1}\binom{n}{a}\binom{n}{b}\cr}
$$
Now the desired inequality for $n+1$ would follow if we have
$$
\frac{a+b+n+1}{a+b+1}\binom{n}{a}\binom{n}{b}\leq \frac{n+1}{a+b+1}\binom{n+1}{a}\binom{n+1}{b}
$$
This is equivalent to
$$
(n+1+a+b)(n+1-a)(n+1-b)\leq(n+1)^3
$$
or
$$
 n+1 \geq \frac{a^2b+b^2a}{a^2+b^2+ab}=b-\frac{b^3}{a^2+ab+b^2}
$$
and this inequality is trivially satisfied since $n+1>b$.
This ends the proof by induction.$\qquad\square$
