Prove that the inequality below holds for all the real numbers $x$, $y$, $z$, such that $1>z>0$:
$(2+z)*(∣x+y+yz∣+∣x+y−yz∣)≥2*z*(|x|+|y|)$
My approach is to consider 8 cases (1 case for $a,b>0$, 3 cases for $a>0$ and $b<0$, etc.). Is there any shorter way to solve this problem?
If $y=0$ so we need to prove that $2(2+z)|x|\geq2z|x|$, which is obvious.
Let $y\neq0$ and $\frac{x}{y}=a$. Hence, we need to prove that: $$(2+z)\left(|z+1+a|+|z-1-a|\right)\geq2z\left(1+|a|\right)$$ If $a\geq0$ so $-1-a\leq z\leq1+a$, which says that we need to prove $$(2+z)(z+1+a-z+1+a)\geq2z(1+a)$$ or $$2+z\geq z$$ If $a\leq0$ so we'll replace $a$ on $-a$ and we need to prove that $$(2+z)\left(|z+1-a|+|z-1+a|\right)\geq2z\left(1+a\right)$$ for $a\geq0$.
Consider two cases.
Hence, $(2+z)\left(|z+1-a|+|z-1+a|\right)\geq(2+z)\left(z+1-a+z-1+a\right)=$
$=2(2+z)z\geq2z\left(1+a\right)$
a) $0<z\leq a-1$.
Hence, we need to prove that $$(2+z)\left(-(z+1-a)+z-1+a\right)\geq2z\left(1+a\right)$$ or $$2(2+z)(1+a)\geq2z(1+a)$$ which is obvious.
b)$z\geq a-1$.
In this case we need to prove that $$(2+z)\left(z+1-a+z-1+a\right)\geq2z\left(1+a\right)$$ or $$2(2+z)z\geq2z(1+a)$$ or $$2+z\geq1+a$$ which is just our case.
Done!