Complex number problem.

Question

For any complex number z, prove that $|Re(z)| + |Im(z)| ≤ (√2)(|z|)$

In my approach, I assumed $z$ to be equal to $x + i y$ which leads me to

LHS = $√x² + √y²$

= $x + y$

And

RHS = $√2 (√(x²+y²))$

= $√(2x² + 2y²)$

Now, the LHS, I know from the given question, has to be positive because it sum of modulus of two quantities, so how do I proceed further proving that LHS ≤ RHS While keeping into consideration the signs of x and y?

Answer 1

Hint $$(a+b)^2\le2(a^2+b^2)$$ for all real numbers $a$ and $b$. There is a one line proof using Cauchy’s inequality and considering the vectors $u=(a,b)$ and $v=(1,1)$. You can also prove it this way: $(a-b)^2=a^2+b^2-2ab\ge 0\Longrightarrow a^2+b^2\ge2ab\Longrightarrow 2(a^2+b^2)\ge a^2+b^2+2ab=(a+b)^2$