Why does the equality would mean that solutions exist?

Question

I was trying to find real numbers p, q and r that satisfy both:
$p^2 + 4q^2 + 9r^2 = 729$ and $ 8p+8q+3r = 243$

I checked the answer, it said that we need to construct an inequality:
$(8p+8q+3r)^2 <=(p^2 + 4q^2+9r^2)(8^2+4^2+1^2)$. The inequality could be derived from scalar product of a and b, by letting $a=\pmatrix{8\\4\\1\\}$ and $b=\pmatrix{p\\2q\\3r}$.

So, if we substitute the values in, and let the equality holds true, for which $a=\lambda b$ we would be able to solve the question. My question is, why do we need to let the equality to hold true. I am very confused.

Thank you so much for your reply.

Answer 1

The root of the discriminant for the quadratic equation in $p$ is $$ \sqrt{ - 784r^2 + 3888r - 729}. $$ This need to be non-negative. Then we have a real solution. This can happen, say, for example with $r=1$ and $$ p=\frac{94+\sqrt{95}}{4}, q=\frac{47-2\sqrt{95}}{8}. $$

Answer 2

The answer tells you to use the Cauchy-Scharz inequality. For two vectors, this reduces to $$|a\cdot b|^2\le|a|^2|b|^2$$ The equality occurs only if $a=\lambda b$. In your case, $a\cdot b=243$, $|a|^2=8^2+4^2+1^2=81$, and $|b|^2=p^2+4q^2+9r^2=729$. You can easily check that $243^2=81\cdot 729$. So in this particular case (for the given numbers), you have equality.

Answer 3

By your work the equality occurs for $$(p,2q,3r)||(8,4,1),$$ which gives that there exist $k$, for which $$(p,q,r)=\left(8k,2k,\frac{k}{3}\right),$$ which gives $$64k+16k+k=243$$ or $k=3$ and $$(p,q,r)=(24,6,1).$$ I like the following way.

By C-S: $$729=p^2+4q^2+9r^2=\frac{(8^2+4^2+1^2)(p^2+4q^2+9r^2)}{81}\geq$$ $$\geq\frac{(8p+8q+3r)^2}{81}=\frac{243^2}{81}=729$$ and the rest is the same.