My Math Forum Prove Inequality

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 February 14th, 2017, 08:22 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 224 Thanks: 26 Prove Inequality Prove inequality for nautral numbers $\displaystyle x_1,x_2,x_3,...,x_n \in N$ $\displaystyle \sqrt{x_1+x_2+...+x_n} \leq \sqrt{x_1}+\sqrt{x_2}+...+\sqrt{x_n}$
 February 14th, 2017, 09:39 AM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,029 Thanks: 420 I would try a proof by weak induction. $\displaystyle Prove\ n \in \mathbb N^+ \implies \sqrt{\sum_{i=1}^nx_i} \le \sum_{i=1}^n \sqrt{x_i}.$ So prove it if n = 1. Simple. $\displaystyle \therefore \exists\ k \in \mathbb N^+\ such\ that\ k \ge 1\ and\ \sqrt{\sum_{i=1}^kx_i} \le \sum_{i=1}^k \sqrt{x_i}.$ Now what?
 February 19th, 2017, 03:52 PM #3 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 600 Thanks: 82 Take two consecutive natural numbers x and x + 1. I don't know how to make the square root symbol, so I'm going to say ^(1/2). Then you have to prove: (x + x + 1)^(1/2) < or = x^(1/2) + (x + 1)^1/2 (2x + 1)^(1/2) < or = x^(1/2) + (x + 1)^1/2 Square both sides. Squaring a monomial is easy. I used first-outer-inner-last (FOIL) to multiply the right binomial by itself. I got: 2x + 1 < or = x + 2(x^2 + x)^(1/2) + x + 1 2x + 1 < or = 2(x^2 + x)^(1/2) + 2x + 1 Subtract (2x + 1) from both sides 0 < or = 2(x^2 + x)^(1/2) If x is natural, 0 < or = 2(x^2 + x)^(1/2) is always true. I hope I did that right. My proof works for any two numbers, such as x and x + 1 or x + 1 and x + 2. I wanted to test if it worked for three or more numbers, such as x, x + 1, and x + 2. I did that proof on paper. The term on the right side gets more complicated the more numbers you use, but the proof works because it ended with 0 < or = to a polynomial. The polynomial will be 0 if x = 0 and the polynomial will be > 0 if x is any other natural number. Last edited by EvanJ; February 19th, 2017 at 04:08 PM.
 February 20th, 2017, 12:12 AM #4 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 For natural numbers $x_{i}$ the following should be obvious $\sqrt{ x_{1} + x_{2} + ... + x_{n} } \leq \sqrt{ x_{1} + x_{2} + ... + x_{n - 1} } + \sqrt{ x_{n}}$ The result follows. Just keep replacing the square root (that has the sums) on the right of the inequality with something obviously BIGGER. Don't touch the left hand side of the inequality. The second replacement is $\sqrt{ x_{1} + x_{2} + ... x_{n - 1 }} \leq \sqrt{ x_{1} + x_{2} + ... + x_{n - 2}} + \sqrt{ x_{n-1} }$ And we get ... $\sqrt{ x_{1} + x_{2} + ... + x_{n}} \leq \sqrt{ x_{1} + x_{2} + ... + x_{n-2}} + \sqrt{ x_{n-1} } + \sqrt{ x_{n}}$ and so on ...

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