Complex analysis proof triangle inequality: 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!
 A: 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.
A: 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 $a^2\le b^2$.
A: 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 $(1)$. 
