Having trouble proving Inequality

Question

I am having trouble proving this inequality: $2ab\leq a^2+b^2$

I can transpose the equation and change around signs. But I am not sure If I need to use k+1 here or just prove the inequality. In discrete math what would my professor be looking for as a proper answer?

A duplicate of this question (among others). It generalizes to more than two variables, and goes by the name AM-GM -inequality. The acronym comes from the fact that the inequality states that the arithmetic mean of a finite collection of positive numbers is $\ge$ than their geometric mean. If you find a better duplicate target, ping me. Dre_Dre: plug in $x=a^2$, $y=b^2$ to arrive at something that looks even more like your inequality. — Jyrki Lahtonen, Jul 21 '16 at 14:16

score 0 · Answer 1 · answered Jul 21 '16 at 14:06

0

HINT: $a^2+b^2-2ab=(a-b)^2\;\;$

answered Jul 21 '16 at 14:06

Brian M. Scott

616,228

$a^2+b^2−2ab=(a−b)^2$, would that be the same as $a^2+b^2−2ab = 0$? – Dre_Dre Jul 21 '16 at 14:10
@Dre_Dre: That would be true if and only if $a=b$. – copper.hat Jul 21 '16 at 14:12
@Dre_Dre: No, it just says that $a^2+b^2-2ab$ is the square of some real number. What do you know about squares? – Brian M. Scott Jul 21 '16 at 14:13
I know that some number squared it that number times itself. When you transpose $2ab$ why does the equation = $(a-b)^2$ ? And not = 0 as in normal equation transposition? – Dre_Dre Jul 21 '16 at 14:21
@Dre_Dre: Multiply out $(a-b)^2$ and see what you get. As for squares, are they ever negative? – Brian M. Scott Jul 21 '16 at 14:22

Having trouble proving Inequality

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