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My professor gave me a hint to subtract one side from both sides and do some algebra so you can group together and combine to make positive numbers.

I am stuck, can anyone help me figure out where to go from here, please? I got

We will prove this directly. We will start with $(a_1b_1+a_2b_2+a_3b_3)^2 \leq ({a_1^2}+{a_2^2}+{a_3^2})({b_1^2}+{b_2^2}+{b_3^2})$ and add the additive inverse of $(a_1b_1+a_2b_2+a_3b_3)^2$ to both sides to get $0 \leq ({a_1^2}+{a_2^2}+{a_3^2})({b_1^2}+{b_2^2}+{b_3^2})-(a_1b_1+a_2b_2+a_3b_3)^2$. We will then expand $({a_1^2}+{a_2^2}+{a_3^2})({b_1^2}+{b_2^2}+{b_3^2})$ to get $0 \leq ({a_1^2b_1^2}+{a_1^2b_2^2}+{a_1^2b_3^2}+{a_2^2b_1^2}+{a_2^2b_2^2}+{a_2^2b_3^2}+{a_3^2b_1^2}+{a_3^2b_2^2}+{a_3^2b_3^2})-(a_1b_1+a_2b_2+a_3b_3)^2$.We will now expand (a_1b_1+a_2b_2+a_3b_3)^2 to get $0 \leq ({a_1^2b_1^2}+{a_1^2b_2^2}+{a_1^2b_3^2}+{a_2^2b_1^2}+{a_2^2b_2^2}+{a_2^2b_3^2}+{a_3^2b_1^2}+{a_3^2b_2^2}+{a_3^2b_3^2})-(a_1^2b_1^2+2a_1a_2b_1b_2+2a_1a_3b_2b_3+a_2^2b_2^2+2a_2a_3b_2b_3+a_3^2b_3^2)$.

Now I do not know how to make this into a positive number so I am stuck. I know that I will have to write the proof backwards once it is completed because you cannot start with what you are trying to prove.

I feel like I might be straying away from the right path, can anyone guide me, please?

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$$0 \leq ({a_1^2b_1^2}+{a_1^2b_2^2}+{a_1^2b_3^2}+{a_2^2b_1^2}+{a_2^2b_2^2}+{a_2^2b_3^2}+{a_3^2b_1^2}+{a_3^2b_2^2}+{a_3^2b_3^2})-(a_1^2b_1^2+2a_1a_2b_1b_2+2a_1a_3b_{\color{red}1}b_3+a_2^2b_2^2+2a_2a_3b_2b_3+a_3^2b_3^2)$$ $$0 \leq {a_1^2b_2^2}+{a_1^2b_3^2}+{a_2^2b_1^2}+{a_2^2b_3^2}+{a_3^2b_1^2}+{a_3^2b_2^2}-2a_1a_2b_1b_2-2a_1a_3b_1b_3-2a_2a_3b_2b_3$$ $$0 \leq ({a_1^2b_2^2}-2a_1a_2b_1b_2+{a_2^2b_1^2})+({a_1^2b_3^2}-2a_1a_3b_1b_3+{a_3^2b_1^2})+({a_2^2b_3^2}-2a_2a_3b_2b_3+{a_3^2b_2^2})$$ $$0 \leq (a_1b_2-a_2b_1)^2+(a_1b_3-a_3b_1)^2+(a_2b_3-a_3b_2)^2$$

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Proof 1. Using the Lagrange's Identity, we have $$(a_1^2+a_2^2+a_3^2 ) ( b_1^2+b_2^2+b_3^2 ) - ( a_1b_1+a_2b_2+a_3b_3 ) ^2$$ $$= ( a_1b_2-a_2b_1 ) ^2+ ( a_1b_3-a_3b_1 ) ^2+ ( a_2b_3-a_3b_2 ) ^2$$ Therefore $$( a_1b_1+a_2b_2+a_3b_3 ) ^2 \leqslant (a_1^2+a_2^2+a_3^2 ) ( b_1^2+b_2^2+b_3^2 ).$$

Proof 2. Let $$f(x) = (a_1^2+a_2^2+a_3^2)x^2-2(a_1b_1+a_2b_2+a_3b_3)x+b_1^2+b_2^2+b_3^2.$$ We have $$f(x) = (xa_1-b_1)^2+(xa_2-b_2)^2+(xa_3-b_3)^2 \geqslant 0,$$ so $$\Delta' = (a_1b_1+a_2b_2+a_3b_3)^2-(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2) \le 0.$$ Therefore $$(a_1b_1+a_2b_2+a_3b_3)^2 \leqslant (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2).$$

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This is a case of the Cauchy-Schwarz Inequality.

The basic inequality for proving Cauchy-Schwarz is a form of the AM-GM: $$ \begin{align} xy &=\frac12\left(x^2+y^2-(x-y)^2\right)\\ &\le\frac12\left(x^2+y^2\right)\tag1 \end{align} $$ We start by assuming that $\sum\limits_{k=1}^na_k^2=\sum\limits_{k=1}^nb_k^2=1$ (i.e. divide $a$ by $|a|$ and $b$ by $|b|$). Then $$ \begin{align} \sum_{k=1}^na_kb_k &\le\frac12\sum_{k=1}^n\left(a_k^2+b_k^2\right)\\ &=1\\ &=\left(\sum_{k=1}^na_k^2\right)^{1/2}\left(\sum_{k=1}^nb_k^2\right)^{1/2}\tag2 \end{align} $$ Inequality $(2)$ is in a form that be scaled in each of $a$ and $b$ (i.e. undo the scaling that was done above), so we can lift the restriction that $\sum\limits_{k=1}^na_k^2=\sum\limits_{k=1}^nb_k^2=1$.

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