Try this problem from ISI-MSQMS 2018 which involves the concept of Inequality and Combinatorics.
Show that $\sqrt{C_{1}}+\sqrt{C_{2}}+\sqrt{C_{3}}+\ldots+\sqrt{C_{n}} \leq 2^{n-1}+\frac{n-1}{2}$ where
$C_k={n\choose k}$
INEQUALITIES
COMBINATORICS
Use Cauchy Schwarz Inequality $\left(\displaystyle\sum_{i} a_{i} b_{i}\right)^{2} \leq\left(\displaystyle\sum_{i} a_{i}^{2}\right)\left(\displaystyle\sum_{i} b_{i}^{2}\right)$
Apply Cauchy Schwarz Inquality in two sets of real numbers ($\sqrt C_1$,$\sqrt C_2$,.....,$\sqrt C_n$)and ($1$,$1$,$1$,......$1$)
($C_1+C_2+$........$+C_n$)($1+1+$......$+1$) $\geq $ ($\sqrt C_1+\sqrt C_2+.........+\sqrt C_n$)
($2^n-1$)$n \geq $ ($\sqrt C_1+\sqrt C_2+$..........$+\sqrt C_n$)$^2$
$\sqrt C_1+\sqrt C_2+$..........$+\sqrt C_n \leq \sqrt n\sqrt (2^n-1)$
The proof is still not done,why don't you try the remaining part yourself?
We know AM $\geq$ GM
i.e
For $n$ positive quantities $a_{1}, a_{2}, \dots, a_{n}$
$$
\frac{a_{1}+a_{2}+\ldots+a_{n}}{n} \geq \sqrt[n]{a_{1} a_{2} \cdot \cdot a_{n}}
$$
with equality if and only if $a_{1}=a_{2}=\ldots=a_{n}$
Now you have all the ingredients,why don't you cook it yourself? I firmly believe that you can cook a food tastier than mine.
$\frac{n+2^n-1}{2} \geq \sqrt n\sqrt {2^n-1}$
Thus,$\sqrt C_1+\sqrt C_2+$........$+\sqrt C_n \leq \frac {n+2^n-1}{2}$
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