# Prove: If $a^2+b^2=1$ and $c^2+d^2=1$, then $ac+bd\le1$

Prove: If $a^2+b^2=1$ and $c^2+d^2=1$, then $ac+bd\le1$

I seem to struggle with this simple proof. All I managed to find is that ac+bd=-4 (which might not even be correct).

• No $ac+bd$ does not equal $-4$. (And even if it did miraculously equal negative four, well, $-4$ is $\le1$...) – anon Jan 3 '15 at 6:01

1st Method

\begin{align}\left(a^2+b^2\right)\left(c^2+d^2\right)=1& \implies (ac+bd)^2+(ad-bc)^2=1\\&\implies (ac+bd)^2\le1\end{align}

2nd Method

\begin{align}\left(a^2+b^2\right)+\left(c^2+d^2\right)=2& \implies \left(a^2+c^2\right)+\left(b^2+d^2\right)=2\\&\implies 2(ac+bd)\le 2\qquad \text{(by A.M.- G.M. Inequality)}\\&\implies (ac+bd)\le 1\end{align}

• An easy way to see this using complex numbers.. $z_1 = a + ib, z_2 = d + ic$, then $|z_1| = |z_2| = 1 \implies |z_1z_2| = 1$ – Aryabhata Jan 3 '15 at 6:10

Alternatively write those numbers in terms of sines and cosines and use the sum formula.

To be more precise, write $a=\sin x,b=\cos x,d=\sin y,c=\cos y$ and note $ac+bd=\sin(x+y)$, in fact also $\cos(x+y)=bc-ad$ so we obtain $$a^2+b^2=1,c^2+d^2=1\implies (ac+bd)^2+(ad-bc)^2=1$$

This can also be seen if we consider complex multiplication. As user Aryabhata noted, we're showing $|z|=|w|=1\implies |zw|=1$. We can show more generally that the complex norm is multiplicative, of course.

• This is a very nice proof! In fact I was about to add that: $a = \sin x, b = \cos x, c = \cos y, d = \sin y$, then $ac + bd = \sin(x+y)$. – Aryabhata Jan 3 '15 at 6:05
• It can even be phrased in terms of complex numbers.. – Aryabhata Jan 3 '15 at 6:09
• Nice answer. But can we really do this not knowing anything about a,b,c,d ? – lsp Jan 3 '15 at 6:09
• @lsp Since $a^2 + b^2 = 1$, the point $(b, a)$ lies on the unit circle. Hence, there exists $x \in \mathbb{R}$ such that $a = \sin x$ and $b = \cos x$. Since $c^2 + d^2 = 1$, the point $(c, d)$ lies on the unit circle. Hence, there exists $y \in \mathbb{R}$ such that $c = \cos y$ and $d = \sin y$. – N. F. Taussig Jan 4 '15 at 15:19

You can use Cauchy Schwarz inequality: $$(ac+bd)^2 \leq (a^2+b^2)\cdot(c^2+d^2)$$ $$(ac+bd) \leq 1$$

• Updated answer. Please refer to the link for Cauchy Schwarz inequality. – lsp Jan 3 '15 at 6:06

$(ac+bd)^2 - (a^2+b^2)(c^2+d^2) = -(ad-bc)^2 \leq 0 \to (ac+bd)^2 \leq 1 \to ...$

• This should not be downvoted. – Suzu Hirose Jan 3 '15 at 6:01

This is asking you to show that the dot product of two $2$-dimensional unit vectors is at most one, and follows immediately if you know about a certain formula related to the dot product: $$ac + bd = \langle a, b \rangle \cdot \langle c, d \rangle = \|\langle a, b \rangle\| \|\langle c, d \rangle\| \cos\theta = \sqrt{a^2 + b^2} \sqrt{c^2 + d^2}\cos\theta = \cos\theta \leq 1$$ (where $\theta$ is the angle between the two vectors). This is basically just the Cauchy-Schwarz inequality: $$ac + bd \leq \sqrt{a^2 + b^2} \sqrt{c^2 + d^2} = 1$$

hint: try to compute $(a-c)^2+(b-d)^2$, and evaluate some inequalities :)