Prove that for any real number $a$, $b$, $c$, $x$, $y$ and $z$, there results $(ax + by + cz)^2 \leq (a^2 + b^2 + c^2)(x^2 + y^2 + z^2)$
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4See Cauchy-Schwarz inequality – Yai0Phah Jul 05 '12 at 13:40
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Another proof is elementary: consider $f(t)=(a^2+b^2+c^2)t^2-2(ax+by+cz)t+(x^2+y^2+z^2)=(at-x)^2+(bt-y)^2+(ct-z)^2\ge0$ whenever $t\in\Bbb R$. – Yai0Phah Jul 05 '12 at 13:41
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Another proof is about Lagrange's identity. – Yai0Phah Jul 05 '12 at 13:43
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2@FrankScience: that's basically the same idea as the proof of Cauchy-Schwarz, isn't it? – Ben Millwood Jul 05 '12 at 13:47
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We observe that \begin{eqnarray*} \text{RHS}&=&(a^2x^2+b^2y^2+c^2z^2)+(a^2y^2+b^2x^2)+(a^2z^2+c^2x^2)+(+b^2z^2+c^2y^2)\\ &=&(a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2czax)+(a^2y^2+b^2x^2-2aybx)+(a^2z^2+c^2x^2-2azcx)+(b^2z^2+c^2y^2-2bzcy)\\ &=&(ax+by+cz)^2+(ay-bx)^2+(az-cx)^2+(bz-cy)^2\\ &\geq&(ax+by+cz)^2\\ &\geq&\text{LHS}. \end{eqnarray*}
blindman
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Let $u=(a,b,c)$ and $v=(x,y,z)$ be two vectors in $\mathbb{R}^3$. We have $$ (\left<u,v\right>)^2=(ax+by+cz)^2 \text{ and } \|u\|^2\|v\|^2=(a^2+b^2+c^2)(x^2+y^2+z^2). $$ Since $(\left<u,v\right>)^2\leq \|u\|^2\|v\|^2$ we have the given inequality.
blindman
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