Prove $(|OP|)+ |PQ|)^2 > |OQ|^2$

Question

I did all the algebra and for some reason I'm getting 0 > $y_2^2$ which is clearly wrong.

Where did I mess up at?

Answer 1

In step 4 from last line, you have both sides are negative so you can't square them. As the rule:$A > B \to A^2 > B^2$ does not apply here and it is true only if $A > B > 0$. So it should be written as: $\sqrt{(x_2 - x_1)^2 + y_2^2} > x_2 - x_1$, and both sides are positive, squaring both sides: $(x_2 - x_1)^2 + y_2^2 > (x_2 - x_1)^2$ and reduces to: $y_2^2 > 0$ which is true, and you are good to go.

Answer 2

From the triangle inequality theorem $$OP + PQ > OQ$$ And this holds also if we square both sides of the inequality, given that the segments lengths are necessarily positive.