Using AM-GM inequality, a+(b/ac)>2(b/c) , b+(c/ab)>2c/a , c+(a/bc)>2a/b, Multiplying these expressions you would automatically get the result.
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Siddharth Kumar
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3 years, 11 months ago

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@Siddharth Kumar
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You're wrong, but the principle is right. \(a+\frac{b}{ac} \geq 2\sqrt{\frac{b}{c}}\). You just forgot the square root. But when you multiply them together, they cancel anyways, leaving a 1 under the square root.
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Bob Krueger
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3 years, 11 months ago

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@Bob Krueger
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Actually I entered the square root but it was not displayed in the answer, don't know why.
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Siddharth Kumar
·
3 years, 11 months ago

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TopNewestUsing AM-GM inequality, a+(b/ac)>2(b/c) , b+(c/ab)>2c/a , c+(a/bc)>2a/b, Multiplying these expressions you would automatically get the result. – Siddharth Kumar · 3 years, 11 months ago

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– Bob Krueger · 3 years, 11 months ago

You're wrong, but the principle is right. \(a+\frac{b}{ac} \geq 2\sqrt{\frac{b}{c}}\). You just forgot the square root. But when you multiply them together, they cancel anyways, leaving a 1 under the square root.Log in to reply

– Siddharth Kumar · 3 years, 11 months ago

Actually I entered the square root but it was not displayed in the answer, don't know why.Log in to reply

– Bob Krueger · 3 years, 11 months ago

Weird. Oh well. It works anyways.Log in to reply

– Hermione Taylor · 3 years, 11 months ago

Get it. Thank you very muchLog in to reply

help me plz we have ab+bc+ca=0 calculate (a+b/c)+(b+c/a)+(c+a/b) – Khaoula Ghayati · 2 years, 7 months ago

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