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I have a problem to prove this inequality $|z_1+z_2|^2 \le (1+|z_1|^2)(1+|z_2|^2)$ $\forall (z_1, z_2)\in \mathbb{C}$.

I tried to take the right hand set and subtract the lfs and after simplification I got this:

$1+(ax)^2+(by)^2 -2(ax+by)+(ay)^2+(bx)^2$ and I couldn't prove thqt this result is positive.

Any help please?

pourjour
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    You need the 1 both in $(ax-1)^2$ and in $(by-1)^2$. Try a square that uses up both $2ax$ and $2by$, and see if $(ay)^2$ and $(bx)^2$ can mop up the remainder – Empy2 Oct 13 '13 at 12:06

4 Answers4

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You can use Cauchy-Schwarz :

$$ |z_1+z_2|^2 \leq (|z_1|+|z_2|)^2 = (1 \times |z_1|+|z_2| \times 1)^2 \leq (1+|z_1|^2)(1+|z_2|^2). $$

Ewan Delanoy
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$$ |a+b|^2\le|a|^2+2|a||b|+|b|^2\le|a|^2+ 1+|a|^2|b|^2+|b|^2=(|a|+1)(|b|+1) $$

Pocho la pantera
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Put $z_k=r_k(\cos\theta_k+i\sin\theta_k),k=1,2$

$(z_1+z_2)^2=r_1^2+r_2^2+2r_1r_2\cos(\theta_1-\theta_2)$

$(1+|z_1|^2)(1+|z_2|^2)=(1+r_1^2)(1+r_2^2)=1+r_1^2+r_2^2+r_1^2r_2^2$

this will be $\ge r_1^2+r_2^2+2r_1r_2\cos(\theta_1-\theta_2)$

$\iff 1+r_1^2r_2^2\ge 2r_1r_2\cos(\theta_1-\theta_2)$

But $1+r_1^2r_2^2\ge 2r_1r_2$ using AM$\ge$ GM

and $r_1r_2\ge r_1r_2\cos(\theta_1-\theta_2)$ as $\cos(\theta_1-\theta_2)\le1$

lab bhattacharjee
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$$ (1+|z_1|^2)(1+|z_2|^2)=|z_1|^2+|z_2|^2+1+|z_1|^2|z_2|^2 $$ Now using AM-GM inequality we have: $$ 1+|z_1|^2|z_2|^2\geq 2|z_1||z_2| $$ Hence $$ (1+|z_1|^2)(1+|z_2|^2)\geq |z_1|^2+|z_2|^2+ 2|z_1||z_2|= (|z_1|+|z_2|)^2 \ge |z_1+z_2|^2 $$

Arash
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