I know that this statement is true, but I cannot figure out a way to actually prove it. $a$ and $b$ are both positive real numbers.
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3You can start from $a^2+b^2-2ab=(a-b)^2>0$ – Emilio Novati May 21 '18 at 21:06
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I got it I just needed to think about it a little more, thanks! – Victor H. May 21 '18 at 21:14
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1@J.G.: the question says $a$ and $b$ are positive real numbers. – Rob Arthan May 21 '18 at 21:16
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1The proof is far easier than any of the algebraic manipulations would suggest. If $a = b$ this is obvious, and if $a > b$, then $ab < a\cdot a < 2a^2 < 2a^2 + 2b^2$. Likewise if $b > a$. – May 21 '18 at 21:19
3 Answers
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$2a^2 + 2b^2 > ab \iff 4a^2 + 4b^2 > 2ab \iff 3a^2 + 3b^2 + (a-b)^2 > 0$ for $a,b$ positive.
bgg1992
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Dividing both sides by $ab$ gives $$2\Big( \frac{a}{b} + \frac{b}{a}\Big) > 1$$ Considering $\mu\equiv\frac{a}{b}$ and cleaning up a bit gives, $$\mu +\mu^{-1} >\frac{1}{2}$$ $a$ and $b$ were positive reals, so $\mu:(0, 1]\cup[1,+\inf)$. Thus, the sum of $\mu$ and it's inverse must be great than one. $\square$
Captain Morgan
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Since $(a-b)^2>0$ for every real $a,b$ not both zero, we have that $$a^2+b^2>2ab>ab/2,$$ as wanted.
Allawonder
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