I'm quite new at calculus, how could I prove:
$$2xy ≤ x^2 + y^2$$
I just can end up in $(xy)^{1/2} ≤ (x+y)/2$. And even though I know that this is true too, I don't know how to prove it either. Thank you in advance :D
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$x^2 + y^2 -2xy \geq0$ – Nosrati Oct 07 '18 at 07:05
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Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. – José Carlos Santos Oct 07 '18 at 07:06
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Observe the fact that for any real $x,y$ that $(x - y)^2 \ge 0$. Is there anything you can do to that inequality to get you to the result you wish to prove? – Decaf-Math Oct 07 '18 at 07:08
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See also: Prove the inequality $|xy|\leq\frac{1}{2}(x^2+y^2)$ and Show that $2 xy < x^2 + y^2$ for $x$ is not equal to $y$ – Martin Sleziak Dec 09 '18 at 07:47
3 Answers
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The trivial inequality $a^2 \geq 0$ is true for any $a$. Im particular, it must be true that
$$ (x-y)^2 \geq 0 $$
But, $(x-y)^2 = x^2 - 2xy + y^2 $. Therefore, the result follows.
James
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$$ 0 \leq (x-y)^2 = x^2 + y^2 - 2xy $$ which yields the result.
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$$x^2 \geq 0$$ $$(a \pm b)^2 \geq 0$$ $$a^2\pm 2ab+b^2 \geq 0$$ $$a^2+b^2 \geq \mp 2ab$$ In particular, using $(a-b)$, we get the following. $$\boxed{a^2+b^2 \geq 2ab}$$
KM101
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