Harder 3U question (Help Needed) (1 Viewer)

visipac · Jul 13, 2012

Give that a² + b² + c²> ab + bc + ca show that:

a) (a + b + c)²> 3(ab + bc + ca)

b) a² + b² + c²> 3xcube root of (a²b²c²)

I need help with b)

asianese · Jul 13, 2012

Using the given, multiply both sides by $(a+b+c)$

$\\(a+b+c)(a^2+b^2+c^2)>(ab+bc+ca)(a+b+c) \\ a^3+ab^2+ac^2+ba^2+b^3+bc^2+ca^2+cb^2+c^3>a^2b+ab^2++abc+abc+b^2c+bc^2+ca^2+abc+c^2a $ cancel out terms $ \\ a^3+b^3+c^3>3abc \\ $ replace $ a^3 $ with $ a^2 $ etc. and the result is obvious $$

visipac · Jul 13, 2012

Thankksss!!

Harder 3U question (Help Needed) (1 Viewer)

visipac

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asianese

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visipac

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