I would like to know how to solve the inequality $$|x^2-y^2|\leq 2x+2y-4xy.$$ I have tried to solve it by myself and searched in the internet, but didn't come up with an answer.
Thanks in advance.
Edit: I forgot to mention that the inequality should be shown for all $x,y\in[0,1]$.
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You've asked how to solve: I'd suggest the answer "by checking all possible cases", for some useful choice of cases. For example, you could try
Case 1: $x^2=y^2$.
Case 1a: $x=y$. Then your inequality is equivalent to $0\leq 4x(1-x)$ with solution....(please fill in!).
Case 1b: $x=-y$. Then your inequality is equivalent to $0\leq4x^2$ with solution...
Case 2: $x^2>y^2$. Then your inequality is equivalent to $x^2-y^2\leq2x+2y-4xy$. If we understand 'solve' in the same way as Wolfram Alpha as in @Amzoti's comment, then it makes sense to try to isolate $y$. We can do this by completing the square and rewriting the inequality in the equivalent form $$(y+1-2x)^2\geq 5x^2-6x+1.\qquad (1)$$
Case 2a: $5x^2-6x+1\leq0$ (corresponding to what restrictions on $x$?). Then (1) gives no further restrictions.
Case 2b: $5x^2-6x+1>0$ (corresponding to what restrictions on $x$?). Then (1) is equivalent to $$y+1-2x <-\sqrt{5x^2-6x+1}\quad\hbox{or}\quad y+1-2x>\sqrt{5x^2-6x+1}.$$ To fully solve Case 2b, combine these restrictions with $x^2>y^2$.
Case 3: $x^2<y^2$. etc...
Note: Your additional condition that $x,y\in[0,1]$ should reduce the number of cases and make this run a bit more smoothly.
I do not understand what "solve" may mean for you. However, did you notice that it is equivalent to $$ \begin{cases} x^2-y^2 \leq 2x+2y-4xy \\ -2x-2y+4xy \leq x^2-y^2 \end{cases} ? $$ Now you have two inequalities without any absolute value, and you can try to "solve" them. Some notion of conic sections will be useful.