$$ac=bd$$

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sqing

#1 h Mar 30, 2025, 2:25 PM

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Let $a,b,c,d$ be reals such that $a^2+b^2=4,c^2+d^2=9$ and $abcd\ge 9.$ Prove that $ac=bd$ Let $a,b,c,d$ be reals such that $a^2+b^2=4,c^2+d^2=9$ and $ad+bc \ge 6.$ Prove that $ac=bd$ Let $a,b,c,d$ be reals such that $a^2+b^2=4,c^2+d^2=9$ and $ab+cd \geq \frac{13}{2}.$ Prove that $ac=bd$

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Primeniyazidayi

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Primeniyazidayi

#3 h Mar 30, 2025, 2:59 PM

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Please somebody turn my solutions to LaTeX(i'm new that's why i can't)
Apply Cauchy-Schwarz and AM-GM to observe that

First one: 36=(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2 \geq 4*abcd \geq 36.Equality conditions gives us the result.

Second one: 36=(a^2+b^2)(d^2+c^2) \geq (ad+bc)^2 \geq 36 and again equality conditions.

Third one: \frac{13}{2} = \frac{9+4}{2} = \frac{a^2+b^2}{2} + \frac{c^2+d^2}{2} \geq ab+cd and again some boring equality cases.
@below thanks

This post has been edited 1 time. Last edited by Primeniyazidayi, Mar 31, 2025, 8:13 AM

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#4 h Mar 30, 2025, 3:04 PM

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Primeniyazidayi wrote:

Please somebody turn my solutions to LaTeX(i'm new that's why i can't)
Apply Cauchy-Schwarz and AM-GM to observe that

First one: $36=(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2 \geq 4 \times abcd \geq 36$ .Equality conditions gives us the result.

Second one: $36=(a^2+b^2)(d^2+c^2) \geq (ad+bc)^2 \geq 36$ and again equality conditions.

Third one: $\frac{13}{2} = \frac{9+4}{2} = \frac{a^2+b^2}{2} + \frac{c^2+d^2}{2} \geq ab+cd$ and again some boring equality cases.

you can put dollar signs around your equations to make it LaTeX

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sqing

#5 h Mar 31, 2025, 3:43 AM

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Thank you.

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