algebra precalculus - How to prove $a^2 + b^2 + c^2 \ge ab
If $\ c> a ,a^2+c^2 \gt 2ac $, because $\ (a-c)^2 \gt 0$ If $\ c>b>a ,c^2+b^2/2+a^2/2 \gt ac+bc $ and $\ b^2/2+a^2/2 \gt ab $, sum of them $\ a^2+b^2+c^2 \gt ab+bc+ac $ If $\ c=b \gt a $ or $\ a=b=c $, it can be solved with same logic.
elementary number theory - Solutions for $a^2+b^2+c^2=d^2 ...
Then $$ a = m^2 + n^2 - p^2 - q^2, $$ $$ b = 2 (mq+np), $$ $$ c = 2(nq -mp), $$ $$ d = m^2 + n^2 + p^2 + q^2 $$ satisfy $$ a^2 + b^2 + c^2 = d^2. $$ This is simply quaternion multiplication. It seems likely that the formula was known to Euler, who wrote out the multiplication formula for sums of four squares, which amounts to quaternion ...
elementary number theory - If $a^2+b^2=c^2$ then $a$ or $b
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Verify that $(a^2 + b^2)(c^2 + d^2)$ = $(ac - bd)^2 + (ad
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Is this correct method to prove that $a^2 + b^2 + c^2 ≥ ab + bc …
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$a^2 + b^2 + c^2 = 1 ,$ then $ab + bc + ca$ gives
2015年2月1日 · $$ (a+b+c)^2 = a^2 + b^2 +c^2 + 2(ab + bc + ca) $$ The LHS of the above identity is a perfect square, hence it is always positive or 0.
proof verification - Does the equation $a^2 + b^2 + c^2 = d^2
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conic sections - Where does $c^2=a^2-b^2$ come from for …
2023年8月29日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
geometry - Alternate proof for $a^2+b^2+c^2\le 9R^2
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algebra precalculus - Where does $b^2 = a^2 - c^2$ come from …
At the level of precalculus, there's a certain derivation of the equation of an ellipse that follows the same path. See this for example. Inevitably at one point in the derivation there's this