Complex analysis proof triangle inequality:

Question

Given: $|z+w|^2=|z|^2+|w|^2+2Re(z\bar w)$ Prove:$|z+w|\leq |z|+|w|$

Work done so far:

Let $z=x+iy$ and $w=a+bi$, then:

$$|x+iy+a+ib|=|z+w|=\sqrt{(x+a)^2+(y+b)^2}$$ $$\sqrt{x^2+y^2}+\sqrt{a^2+b^2}=|z|+|w|$$ Squaring it I get, $$x^2+y^2+2|z||w|+a^2+y^2$$

After this I am lost, any idea how to proceed or if I am doing it wrong, what is the right way.

Any help is appreciated!

Answer 1

You want to show that $$\tag1 \sqrt{(x+a)^2+(y+b)^2}\leq \sqrt{a^2+b^2}+\sqrt{x^2+y^2}. $$ If you look at the squares, that would be $$\tag2 {(x+a)^2+(y+b)^2}\leq a^2+b^2+x^2+y^2+2\sqrt{a^2+b^2}\,\sqrt{x^2+y^2}, $$ which reduces to $$\tag3 2ax+2yb\leq 2\sqrt{a^2+b^2}\,\sqrt{x^2+y^2}, $$ and squaring again this is (after cancelling terms), $$\tag4 0\leq 4a^2y^2+4b^2x^2. $$

is obviously true. So, if we start from $(4)$, we add $4a^2x^2+4b^2y^2$ to both sides, to get $$ (2ax+2by)^2\leq 4(a^2+b^2)(x^2+y^2). $$ Taking square root (everything is positive), we get $$ 2ax+2by\leq|2ax+2by|\leq2\sqrt{a^2+b^2}\,\sqrt{x^2+y^2}. $$ Now add $a^2+b^2+x^2+y^2$ to both sides, to get $(2)$, and taking square roots again we get $(1)$.

Answer 2

You just need to show that $$ \operatorname{Re}(z\bar{w})\le |z||w| \tag{*} $$ because if this is true, then $$ |z+w|^2=|z|^2+|w|^2+2\operatorname{Re}(z\bar{w})\le |z|^2+|w|^2+2|z||w|=(|z|+|w|)^2 $$ Now (*) is obvious if $\operatorname{Re}(z\bar{w})<0$. Suppose it is $\ge0$. Then, with your notation, you have to prove that $$ (xa+yb)\le\sqrt{x^2+y^2}\sqrt{a^2+b^2} $$ Square both sides and conclude.

Alternatively, observe that $\lvert\operatorname{Re}{z}\rvert\le\lvert z\rvert$; this will lead to a shorter proof.

Answer 3

HINT: Please do not write out real and imaginary parts. Just use the inequality you were given. Compare $|z+w|^2$ and $(|z|+|w|)^2$, and recall what you know (or can prove) if $a,b\ge 0$ and $a^2\le b^2$.