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26/07/2017 at 08:54

Given \(a,b,c>0\), prove that:
\(a^3+b^3+c^3\ge a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}\).

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Lufans 26/07/2017 at 15:01

Set \(A=a^3+b^3+c^3\); \(B=a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}\)

Apply AM-GM inequality for 2 positive numbers

\(a^3+abc\ge2\sqrt{a^3.abc}=2a^2\sqrt{bc}\)

\(b^3+abc\ge2\sqrt{b^3.abc}=2b^2\sqrt{ac}\)

\(c^3+abc\ge2\sqrt{c^3.abc}=2c^2\sqrt{ab}\)

Thence inferred \(A+3abc\ge2B\)

Need proof: \(3abc\le A\) it mean \(3abc\le a^3+b^3+c^3\), this is always true because of the AM-GM inequality we have \(a^3+b^3+c^3\ge3\sqrt[3]{a^3.b^3.c^3}=3abc\)

Equal sign occurs when and only if a = b = c
Faded 28/01/2018 at 21:28

Set $A = a^{3} + b^{3} + c^{3}$ ; $B = a^{2} \sqrt{b c} + b^{2} \sqrt{a c} + c^{2} \sqrt{a b}$

Apply AM-GM inequality for 2 positive numbers

$a^{3} + a b c \geq 2 \sqrt{a^{3} . a b c} = 2 a^{2} \sqrt{b c}$

$b^{3} + a b c \geq 2 \sqrt{b^{3} . a b c} = 2 b^{2} \sqrt{a c}$

$c^{3} + a b c \geq 2 \sqrt{c^{3} . a b c} = 2 c^{2} \sqrt{a b}$

Thence inferred $A + 3 a b c \geq 2 B$

Need proof: $3 a b c \leq A$

it mean $3 a b c \leq a^{3} + b^{3} + c^{3}$ , this is always true because of the AM-GM inequality we have $a^{3} + b^{3} + c^{3} \geq 3 \sqrt[3]{a^{3} . b^{3} . c^{3}} = 3 a b c$

Equal sign occurs when and only if a = b = c

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