Cauchy-Schwarz inequality for i=2 ; the quickest proof possible

Question

I've already found some proofs of this inequality: $$\sum_{i=1}^n a_i b_i \le \sqrt {\biggl(\sum_{i=1}^n a_i^2\biggr)\biggl(\sum_{i=1}^n b_i^2\biggr)}$$ (using dot product) but they were far too long. I need only a proof for i=2, meaning I need to show that $$ a_1 b_1 + a_2 b_2 \leq \sqrt{(a_1^2+a_2^2)(b_1^2+b_2^2)} $$ or, simillary: $$ a c + b d \leq \sqrt{(a^2+b^2)(c^2+d^2)} $$ I rewrote the formula to suggest it would be the best if it was possible to prove that just by using simple algebra. I'm not lazy but the proof is supposed to be only a lemma so I basically don't have enough time to understand and memorize one of those longer prooves. I've just started studying in high school (not even have learnt about calculus that accurately, just basic rules, etc) so please, be forgiving.

So my question is: how to prove that equation as quickest and easiest as possible?

Thanks in advance,

somnock

Answer 1

Sketch the triangle with vertices at $(0,0)$, $(a,b)$, and $(c,d)$, which has side lengths $\sqrt{a^2+b^2}$, $\sqrt{c^2+d^2}$, and $\sqrt{(a-c)^2+(b-d)^2}$, with the latter side being opposite the angle I'll call $\theta$. By the law of cosines,

$$(a-c)^2+(b-d)^2 = a^2+b^2 + c^2 + d^ 2 -2\sqrt{a^2+b^2}\sqrt{c^2+d^2}\cos(\theta),$$ which reduces to $$ac+bd = \sqrt{a^2+b^2}\sqrt{c^2+d^2}\cos(\theta).$$

The result follows from taking absolute values, because $|\cos(\theta)|\leq 1$.

Answer 2

For the case of $i = 2$, note that it is enough to show that $$ (1 + ab) \leq \sqrt{(1 + a^2)(1 + b^2)}.$$ You get the desired result if you put $a = \frac{a_1}{a_2}$ and $b = \frac{b_1}{b_2}$ with $a_2 \ne 0, b_2 \ne 0.$ $$(1+a^2)(1+b^2) = (1 + a^2b^2 + a^2 + b^2) \geq (1 + a^2b^2 + 2ab).$$ Therefore we have $$(1 + a^2)(1+b^2) \geq (1 + ab)^2.$$ Taking square root on both sides we get $$(1 + ab) \leq \sqrt{(1 + a^2)(1 + b^2)}.$$ The answer actually follows from @DanielFischer's comment. It is just that it is easier to see the result with the normalization.

Answer 3

Consider the complex numbers $z=x_1+iy_1$, $w=x_2-iy_2$. The norm of the product is the product of the norms, or $$|z|^2|w|^2=(x_1^2+x_2^2)(y_1^2+y_2^2)=(x_1x_2+y_1y_2)^2+(y_1x_2-x_1y_2)^2=|zw|^2$$

Since $(y_1x_2-x_1y_2)^2$ is positive, this says that $$(x_1^2+x_2^2)(y_1^2+y_2^2)\geqslant (x_1x_2+y_1y_2)^2$$

unless $y_1x_2=x_1y_2$.

Answer 4

By brute force algebra, we have $$ (a^2 + b^2)(c^2 + d^2) - (ac +bd)^2 = a^2d^2 - 2abcd + b^2c^2 = (ad - bc)^2 \ge 0 $$ So $$ (ac + bd)^2 \le (a^2 + b^2)(c^2 + d^2) $$ and the desired result follows immediately (as Daniel Fischer said).

Answer 5

I think the best proof is to consider the polynomial of $x$

$$(a_1x+b_1)^2 + (a_2x+b_2)^2 = (a_1^2+a_2^2)x^2 + 2(a_1b_1+a_2b_2)x +(b_1^2+b_2^2)$$

From the left hand side, the polynomial is never less than zero, so the discriminant is never greater than zero:

$$2^2(a_1b_1+a_2b_2)^2 - 4(a_1^2+a_2^2)(b_1^2+b_2^2)\le 0$$

This is the proof I learnt, and it works for general natural number $n$ and real $a$'s and $b$'s. Equality holds when all $\frac{a_i}{b_i}$ are equal (or all $b_i = 0$).