How to show that $\left(\sqrt{\binom{n}{1}}+\sqrt{\binom{n}{2}}+...+\sqrt{\binom{n}{n}}\right)^2≤n(2^n-1)$? I want to show that $\left(\sqrt{\binom{n}{1}}+\sqrt{\binom{n}{2}}+...+\sqrt{\binom{n}{n}}\right)^2≤n(2^n-1)$.
My attempt:
We know that $\sqrt{ab}≤\frac{a+b}{2}$ because $(\sqrt{a}-\sqrt{b})^2≥0$ and that $\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+...+\binom{n}{n}=2^n$ by expanding $(1+x)^n$ and substituting $x=1$.
Next, I re-wrote $\left(\sqrt{\binom{n}{1}}+\sqrt{\binom{n}{2}}+...+\sqrt{\binom{n}{n}}\right)^2$ as $\binom{n}{1}+\binom{n}{2}+...+\binom{n}{1}+\sum_{0<i<j}\sqrt{\binom{n}{i}\binom{n}{j}}$
$≤2^n-1+\sum_{0<i<j}\frac{\binom{n}{i}+\binom{n}{j}}{2}$ using the two term $A_m G_m$ inequality.
$=2^n-1+\frac{1}{2}\left(\binom{n}{1}+\binom{n}{2}+\binom{n}{1}+\binom{n}{3}+...+\binom{n}{1}+\binom{n}{n}+\binom{n}{2}+\binom{n}{3}+\binom{n}{2}+\binom{n}{4}+...+\binom{n}{n-1}+\binom{n}{n}\right)$
$=2^n-1+\frac{n}{2}\left(\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+...+\binom{n}{n}\right)$
$=2^n-1+\frac{n}{2}(2^n-1)$
$=\frac{n+2}{2}(2^n-1)$, which is clearly not the result.
Could someone point out the mistake in my working and help me finish the proof?

There is an error in your calculation. You should have
$$
 \left(\sqrt{\binom{n}{1}}+\sqrt{\binom{n}{2}}+...+\sqrt{\binom{n}{n}}\right)^2 = \sum_{i=1}^n \binom ni + 2 \sum_{0<i<j}\sqrt{\binom{n}{i}\binom{n}{j}} \\
\le
\sum_{i=1}^n \binom ni  + \sum_{0<i<j} \left(\binom{n}{i} + \binom{n}{j}\right)\, ,
$$
note the factor $2$ which is missing in your equation. The right-hand side is equal to
$$
 \frac 12 \sum_{i=1}^n\sum_{j=1}^n \left(\binom{n}{i} + \binom{n}{j}\right) = n \sum_{i=1}^n \binom{n}{i} = n (2^n-1) \, .
$$
