When does the inequality $|x+y| \le |x|+|y|$ not hold? I was trying to find the least value of $|\sin x|+ |\cos x|$ and applied the above inequality as :
$|\sin x + \cos x| \le |\sin x| + |\cos x|\\
\Longrightarrow  \sqrt{2} \le |\sin x| + |\cos x|$
But the range of the given function is $[1,\sqrt2]$. Any idea what might have went wrong here?
 A: It doesn't hold iff $x$ and $y$ have the same sign. As for your problem remember that if $a\leq 1$ then $a^2\leq a$ so $$ |\sin x|+|\cos x|\geq |\sin x|^2+|\cos x|^2=1$$
and for upper bound (remember $(a+b)^2\leq 2(a^2+b^2)$): $$|\sin x|+|\cos x| \leq \sqrt{2(|\sin x|^2+|\cos x|^2)}= \sqrt{2}$$
A: $|\sin x + \cos x| \le |\sin x| + |\cos x|$ is correct.
But, for example, if $x=0$ then both sides are $1$; thus we cannot conclude
$\sqrt{2} \le |\sin x| + |\cos x|$.
A: Let $\begin{cases}f(x)=\sin(x)+\cos(x)\\g(x)=|\sin(x)|+|\cos(x)|\end{cases}$
Notice that $g(x+\frac{\pi}2)=|\cos(x)|+|\sin(x)|=g(x)$ so
$g$ is periodic of period $\frac{\pi}{2}$ and we can study it on $[0,\frac{\pi}2]$.
But on this interval both $\sin$ and $\cos$ are positive, so we can get rid of absolute values and $g(x)=f(x)$ on $[0,\frac{\pi}2]$.
You now can use the addition formulas to transform $f$ to $f(x)=\sqrt{2}\sin(x+\frac{\pi}4)$ and see that $|f(x)|$ has values between $0$ and $\sqrt{2}$ while $g(x)$ has values between $1$ and $\sqrt{2}$.
