How to proceed further with the following inequality problem that can be used by the arithmetic and geometric inequality theorem

Question

I am an Advanced level math student and we were given to prove the following inequality for homework,

If $a,b,c,p,q,r$ are positive real numbers show that, $(a^2+b^2+c^2)(p^2+q^2+r^2)\geq(ap+bq+cr)^2$

I actually gave this problem a try and below is what I have done so far,

By using the arithmetic and geometric inequality theorem we can obtain the following three results,

$\frac{a^2+b^2+c^2}{3}\geq (\sqrt[3]{abc})^2------>(1)$
$\frac{p^2+q^2+r^2}{3}\geq (\sqrt[3]{pqr})^2------>(2)$
$\frac{ap+bq+cr}{3}\geq \sqrt[3]{apbqcr}------>(3)$

From $(1)\times(2)$ we get,
$\frac{(a^2+b^2+c^2)(p^2+q^2+r^2)}{9}\ge(\sqrt[3]{abcpqr})^2------>(4)$

By obtaining the square of the relationship (3), we get,
$\frac{(ap+bq+cr)^2}{9}\geq (\sqrt[3]{apbqcr})^2------>(5)$

From $\frac{(4)}{(5)}$ we get,
$\frac{(a^2+b^2+c^2)(p^2+q^2+r^2)}{(ap+bq+cr)^2}\ge1\Rightarrow (a^2+b^2+c^2)(p^2+q^2+r^2) \ge (ap+bq+cr)^2$

Here, what I do not understand is, I know that we can not divide two inequalities by each other. So I know there must be a mistake in my proof when I did the $\frac{(4)}{(5)}$. So how can I get the correct proof without dividing the two inequalities? What is the correct way to obtain the given relationship?

I would be really grateful if you can answer my question. Thanks.

Answer 1

Can't this be shown directly?

$(a^2+b^2+c^2)(p^2+q^2+r^2) ≥ (ap+bq+cr)^2$

\begin{array}{ccccccccccccc} a^2p^2 &+ &a^2q^2 &+ &a^2r^2 &+ &≥ &a^2p^2 &+ &abpq &+ &acpr &+\\ b^2p^2 &+ &b^2q^2 &+ &b^2r^2 &+ & &abpq &+ &b^2q^2 &+ &bcqr &+\\ c^2p^2 &+ &c^2q^2 &+ &c^2r^2 & & &acpr &+ &bcqr &+ &c^2r^2 & \newline \newline & &a^2q^2 &+ &a^2r^2 &+ &≥ & & &abpq &+ &acpr &+\\ b^2p^2 & & &+ &b^2r^2 &+ & &abpq & & &+ &bcqr &+\\ c^2p^2 &+ &c^2q^2 & & & & &acpr &+ &bcqr & & &\\ \newline & &a^2q^2 &+ &a^2r^2 &+ &≥ &2abpq &+\\ b^2p^2 & & &+ &b^2r^2 & & &2acpr &+\\ c^2p^2 &+ &c^2q^2 & & & & &2bcqr\\ \end{array}

\begin{array}{ccccc} a^2q^2 &+ &b^2p^2 &≥ &2abpq\\ a^2r^2 &+ &c^2p^2 &≥ &2acpr\\ b^2r^2 &+ &c^2q^2 &≥ &2bcqr\\ \end{array}

\begin{equation}\begin{aligned} q^2/b^2 + p^2/a^2 &≥ 2pq/ab\\ q' &= q/b\\ p' &= p/a\\ q'^2 + p'^2 &≥ 2p'q'\\ q'^2 - 2p'q' + p'^2 &≥ 0\\ (q' - p')^2 &≥ 0\\ \end{aligned} \end{equation}

Answer 2

For any two $3$-dimensional vectors $\textbf{a}$ and $\textbf{b}$, we have:

$$\textbf{a}\cdot\textbf{b} = a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3} = \vert\vert \textbf{a}\vert\vert\ \vert\vert\textbf{b}\vert\vert\cos(\theta)$$

Where $\theta$ is the angle between the vectors. Note that $\cos(\theta) \leq 1$.

Consider the dot product of two vectors $\textbf{p} = (a, b, c)$ and $\textbf{q} = (x, y, z)$. We have

$$\textbf{p}\cdot\textbf{q}= ax + by + cz = \vert\vert \textbf{p}\vert\vert\ \vert\vert\textbf{q}\vert\vert\cos(\theta)\leq \vert\vert\textbf{p}\vert\vert\ \vert\vert\textbf{q}\vert\vert = (a^{2} + b^{2} + c^{2})(x^{2} + y^{2} + z^{2})$$

This proves the inequality. It also gives us the equality case, namely $\cos(\theta) = 1$, or $\theta = 0$. This is equivalent to the vectors being parallel and pointing in the same direction, or:

$$\frac{a}{x} = \frac{b}{y} = \frac{c}{z} \geq 0$$

This is called the Cauchy - Schwarz Inequality, and it can be generalized to any number of arguments with this dot product method.

Answer 3

Use Cauchy-Scwartz inequality: $$(ax-p)^2+(bx-q)^2+(cx-r)^2 \ge 0, \forall x \in \Re$$ $$\implies (a^2+b^2+c^2)x^2-2(ap+bq+cq)x+(p^2+q^2+r^2) \ge 0 \forall x \in R$$ If $Ax^2+Bx+C\ge 0, \forall x \in R$, then the discriminant $A>0, B^2-4AC \le 0$, this gives $$(ap+bq+cr)^2 \le (a^2+b^2+c^2)(p^2+q^2+r^2).$$ Equality, holds when $a/p=b/q=c/r=k, k \in R$