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Problem I can't solve

"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

Note by Jan Sadowski
2 years, 8 months ago

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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.

- 2 years, 8 months ago

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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, $$x =2014$$.

- 2 years, 8 months ago

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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.

- 2 years, 8 months ago

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