If $a$ and $b$ are any two real numbers with opposite signs, which of the following is the greatest?

(a - b)^{2}

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(| a | - | b |)^{2}

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| a^{2} - b^{2} |

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a^{2} + b^{2}

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Solution

The correct option is A $(a - b)^{2}$
$(a - b)^{2} = a^{2} + b^{2} - 2 a b$
as a and b are of oppsite sign ab<0 and -2ab>0,it means $(a - b)^{2} > a^{2} + b^{2} - 2 | a | | b | = (| a | - | b |)^{2} > (| a 2 + b^{2} |) > (| a^{2} - b^{2} |)$

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z_{1}

z_{2}

{| (a z_{1} - b z_{2}) |}_{1}^{2} + {| b z_{1} + a z_{2} |}_{1}^{2} =

a, b, c

a^{2} = b^{2} + c^{2} - 2 b c \sqrt{1 - a^{2}}, b^{2} = c^{2} + a^{2} - 2 c a \sqrt{1 - b^{2}}, c^{2} = a^{2} + b^{2} - 2 a b \sqrt{1 - c^{2}}

a = c \sqrt{1 - b^{2}} + b \sqrt{1 - c^{2}}

For any two complex numbers $z_{1} z_{2}$ and any two real numbers a and b $| a z_{1} - b z_{2} |^{2} + | b z_{1} + a z_{2} |^{2}$ =

z_{1}, z_{2}

a

b

| a z_{1} - b z_{2} |^{2} + | b z_{1} + a z_{2} |^{2} =

\frac{b^{2} + c^{2}}{b + c} + \frac{c^{2} + a^{2}}{c + a} + \frac{a^{2} + b^{2}}{a + b} \geq a + b + c

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