inequality $|x + 3\eta^2| \geq |x| + |\eta|^2$

Question

I'm learning the basics of Fourier Analysis. Reading a book, I've found this inequality: " If $x\geq -3$ and $|\eta | >2$ we have $|x + 3\eta^2| \geq |x| + |\eta|^2$. I've tried to solve this inequality applying triangle one but I failed. Can you help me, please?

Answer 1

By the comments of @Gregory, it suffices to consider $-3<x<0$. In that case $0<|x|=-x<3$.

Since $\eta^{2}>4>|x|$, adding $\eta^{2}-|x|$ to both sides we get $$2\eta^{2}>2|x|\Longrightarrow 3\eta^{2}-|x|>\eta^{2}+|x|.$$ Observe that $$3\eta^{2}-|x|=|3\eta^{2}+x|,$$ when $x<0$.

Answer 2

Under your hypothesis, the inequality you want to prove is equivalent to $$x+3\eta^2\ge|x|+\eta^2$$ i.e. $$2\eta^2\ge|x|-x,$$ which is obvious if $x\ge0$, and is a consequence of $x+\eta^2>0$ if $x\le0$.