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Let $a$ and $b$ be two real, positive numbers. Is it possible to prove $$a^2+b^2 \ge 2ab$$ using the Triangle Inequality?

This was suggested to me as a proof method but I have been unsuccessful so far.

Martin Sleziak
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flow2k
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    Would you like to think of a triangle inequality solution, or is the standard method fine? – Ninad Munshi Dec 29 '19 at 07:36
  • if a, b are both positive then |a|+|b|=a+b=|a+b| though – Divide1918 Dec 29 '19 at 07:38
  • @NinadMunshi I'd be interested in other proof methods too. My first proof was to use the fact $a^2+b^2-2ab=(a-b)^2$. Is this the method you had in mind? – flow2k Dec 29 '19 at 07:49
  • @NinadMunshi I'd guess that there are plenty of questions on the site which contain proof by other methods, such as: https://math.stackexchange.com/q/320244, https://math.stackexchange.com/q/470221, https://math.stackexchange.com/q/943994, https://math.stackexchange.com/q/64881, https://math.stackexchange.com/q/543253, https://math.stackexchange.com/q/1493711, etc. – Martin Sleziak Dec 29 '19 at 08:03
  • @flow2k I wonder where the suggestion to use the triangle inequality in this proof came from. (It might be useful to know especially if it came from some text. It would be also a reasonable way to provide context.) – Martin Sleziak Dec 29 '19 at 08:12
  • @MartinSleziak Got it - I didn't add additional references, as it came from a discussion I just had with a friend, when we were discussing the magnitude of $(x^2+y^2)/2$ vs that of $((x+y)/2)^2$. – flow2k Dec 29 '19 at 08:22

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By the triangle inequality $$|a-b|+|b-a|\geq|a-b+b-a|=0.$$ Thus, $$|a-b|\geq0$$ or $$(a-b)^2\geq0$$ or $$a^2+b^2\geq2ab.$$

Michael Rozenberg
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    Why did you need the triangle inequality to deduce that $(a-b)^2 \geq 0$? – Mustafa Said Dec 29 '19 at 09:03
  • @Mustafa Said Because the topic starter wants a proof by the triangle inequality. See please the starting problem. – Michael Rozenberg Dec 29 '19 at 09:05
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    The prroof seems a little strange because you could have just started with $(a-b)^2 \geq 0$ which obviously holds for all real numbers. – Mustafa Said Dec 29 '19 at 09:07
  • I agree with you, @Mustafa but the topic starter wants a proof by using a triangle inequality. I proved that $|a-b|\geq0$, by the triangle inequality, but it's obvious without this way. – Michael Rozenberg Dec 29 '19 at 09:11