# 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

$$≤2^n-1+\sum_{0 using the two term $$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?

## 2 Answers

Hint: Use the CS inequality: $$(1\cdot x_1+1\cdot x_2+\cdots+1\cdot x_n)^2 \le (1+1+\cdots+1)(x_1^2+x_2^2+\cdots+x_n^2)$$, with $$x_k = \sqrt{\binom{n}{k}}$$

• Seems more like a solution than a hint to me. If you want to phrase it as a hint (which I'm generally in favor of), maybe hide the inequality? Feb 22, 2022 at 14:08

There is an error in your calculation. You should have $$\left(\sqrt{\binom{n}{1}}+\sqrt{\binom{n}{2}}+...+\sqrt{\binom{n}{n}}\right)^2 = \sum_{i=1}^n \binom ni + 2 \sum_{0 note the factor $$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) \, .$$