Problem I can't solve

"Determine which of the numbers: 2011+2013+2015+2017 \sqrt { 2011 } +\sqrt { 2013 } +\sqrt { 2015 } +\sqrt { 2017 } and 420144\sqrt { 2014 } is greater. " Do you have any ideas? :D

#Algebra #Problem #Roots #Greater #Determine

Note by Jan Sadowski
6 years, 1 month ago

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Comments

The second number, 420144\sqrt{2014} is larger.

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

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

(x−3−x)+(x+3−x)≥0(x−3+x+3)−2x≥0(x−3+x+3)2−(2x)2≥02x+2x2−9≥4x2x+2x2−9≥4x2x2−9≥2x4(x2−9)≥4x2−36≥0 Which is absurd \begin{aligned} ( \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{aligned}

Thus (x−3−x)+(x+3−x)<0 ( \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=2014x =2014.

Pi Han Goh - 6 years, 1 month ago

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

Pi Han Goh - 6 years, 1 month ago

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

For ∣a∣<1,∣b∣<1 \mid a \mid < 1, \mid b \mid < 1 .

Binomal expansion: (1±c)n=1±nc+n(n−1)2c2+… (1\pm c)^n = 1 \pm nc + \frac{n(n-1)}{2} c^2 + \ldots .

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

Pi Han Goh - 6 years, 1 month ago
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