"Determine which of the numbers: \( \sqrt { 2011 } +\sqrt { 2013 } +\sqrt { 2015 } +\sqrt { 2017 } \) and \(4\sqrt { 2014 }\) is greater. "
Do you have any ideas? :D

We can also apply calculus. Define \(f(x) = \sqrt x \), then \(f'(x) > 0, f''(x) < 0 \) for \(x> 3 \). Then the difference between \( \sqrt{x-3} - \sqrt{x} \) and \( \sqrt{x+3} - \sqrt{x} \) is strictly negative. Similarly the difference between \( \sqrt{x-1} - \sqrt{x} \) and \( \sqrt{x+1} - \sqrt{x} \) is strictly negative. Add them up shows that \( ( \sqrt{x-3} - \sqrt{x} ) + ( \sqrt{x+3} - \sqrt{x} )+ ( \sqrt{x-1} - \sqrt{x} ) + ( \sqrt{x+1} - \sqrt{x} ) < 0 \). In this case, \(x=2014\). So the second number is larger.

Thus \( ( \sqrt{x-3} - \sqrt{x} ) + ( \sqrt{x+3} - \sqrt{x} ) < 0 \). Using an analogous argument, we can see that the sum of the other two terms is negative as well.

Thus the inequality in question holds true. In this case, \(x =2014\).

Taking \(a = \frac 3 x, b= \frac 1 x, n = \frac 12 \) shows that the expression \( \bigg [ (1 + a)^n + (1-a)^n + (1+b)^n + (1-b)^n - 4
\bigg ] \) is negative. With the substitution of \(x=2014\) proves that the second number is larger.

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TopNewestWe can also apply calculus. Define \(f(x) = \sqrt x \), then \(f'(x) > 0, f''(x) < 0 \) for \(x> 3 \). Then the difference between \( \sqrt{x-3} - \sqrt{x} \) and \( \sqrt{x+3} - \sqrt{x} \) is strictly negative. Similarly the difference between \( \sqrt{x-1} - \sqrt{x} \) and \( \sqrt{x+1} - \sqrt{x} \) is strictly negative. Add them up shows that \( ( \sqrt{x-3} - \sqrt{x} ) + ( \sqrt{x+3} - \sqrt{x} )+ ( \sqrt{x-1} - \sqrt{x} ) + ( \sqrt{x+1} - \sqrt{x} ) < 0 \). In this case, \(x=2014\). So the second number is larger.

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The second number, \(4\sqrt{2014}\) is larger.

We need to prove that \( ( \sqrt{x-3} - \sqrt{x} ) + ( \sqrt{x+3} - \sqrt{x} )+ ( \sqrt{x-1} - \sqrt{x} ) + ( \sqrt{x+1} - \sqrt{x} ) < 0 \) for all \(x > 3 \).

We want to prove that the sum of the first two terms is negative by contradiction. Suppose it's non-negative, then

\[ \begin{eqnarray} ( \sqrt{x-3} - \sqrt{x} ) + ( \sqrt{x+3} - \sqrt{x} ) & \geq & 0 \\ ( \sqrt{x-3} + \sqrt{x+3} ) - 2 \sqrt x & \geq & 0 \\ \sqrt{( \sqrt{x-3} + \sqrt{x+3} )^2} - \sqrt{(2 \sqrt x)^2} & \geq & 0 \\ \sqrt{2x + 2\sqrt{x^2-9}} & \geq & \sqrt{4x} \\ 2x + 2\sqrt{x^2-9} & \geq & 4x \\ 2\sqrt{x^2-9} & \geq & 2x \\ 4(x^2-9) & \geq & 4x^2 \\ -36 & \geq & 0 \text{ Which is absurd} \\ \end{eqnarray}\]

Thus \( ( \sqrt{x-3} - \sqrt{x} ) + ( \sqrt{x+3} - \sqrt{x} ) < 0 \). Using an analogous argument, we can see that the sum of the other two terms is negative as well.

Thus the inequality in question holds true. In this case, \(x =2014\).

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Alternatively, one could solve this by binomial expansion. For \(x > 3 \). Consider the expression \[ \sqrt{x} \cdot \bigg [ (1 + a)^n + (1-a)^n + (1+b)^n + (1-b)^n - 4 \bigg ] \]

For \( \mid a \mid < 1, \mid b \mid < 1 \).

Binomal expansion: \( (1\pm c)^n = 1 \pm nc + \frac{n(n-1)}{2} c^2 + \ldots \).

Taking \(a = \frac 3 x, b= \frac 1 x, n = \frac 12 \) shows that the expression \( \bigg [ (1 + a)^n + (1-a)^n + (1+b)^n + (1-b)^n - 4 \bigg ] \) is negative. With the substitution of \(x=2014\) proves that the second number is larger.

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