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Question Number 122860 by greg_ed last updated on 20/Nov/20

	Answered by MJS_new last updated on 20/Nov/20
	$both sides ≥0 ⇒ squaring allowed x^2 +y^2 +2∣xy∣≤2(x^2 +y^2 )+2∣x^2 −y^2 ∣ −x^2 +2∣xy∣−y^2 ≤2∣x^2 −y^2 ∣ now xy is either <0 or ≥0 {: ((−(x+y)^2 )),((−(x−y)^2 )) }≤2∣x^2 −y^2 ∣ true since lhs≤0 and rhs≥0$
	$both sides ⩾ 0 \Rightarrow squaring allowed$ $x^{2} + y^{2} + 2 ∣ x y ∣⩽ 2 (x^{2} + y^{2}) + 2 ∣ x^{2} - y^{2} ∣$ $- x^{2} + 2 ∣ x y ∣ - y^{2} ⩽ 2 ∣ x^{2} - y^{2} ∣$ $now x y is either < 0 or ⩾ 0$ $\begin{matrix} - {(x + y)}^{2} \\ - {(x - y)}^{2} \end{matrix}} ⩽ 2 ∣ x^{2} - y^{2} ∣ true since lhs ⩽ 0 and rhs ⩾ 0$
	Commented by greg_ed last updated on 20/Nov/20

	$thanks .$ $but, i need another way!$
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