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Help me prove $a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\geq6abc$

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What are $a,b,c$? Real numbers? –  Zev Chonoles Jul 15 '12 at 14:47
    
Hint: $1 + b^2 \geq 2b$, $1 + c^2 \geq 2c$,... –  Zarrax Jul 15 '12 at 14:52
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Help me prove + asked 31 mins ago + answered 26 mins ago + 0% accept rate = winning combo. –  Did Jul 15 '12 at 15:18

2 Answers 2

up vote 3 down vote accepted

$a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\geq6abc$

Since

$(a-bc)^2\geq 0$,

$(b-ac)^2\geq 0$,

$(c-ab)^2\geq 0$,

then

$a^2+b^2c^2\geq 2abc$,

$b^2+a^2c^2\geq 2abc$,

$c^2+a^2b^2\geq 2abc$.

By of collectted through for through three inequalities last will be obtained

$a^2+b^2c^2+b^2+a^2c^2+c^2+a^2b^2\geq6abc$,

or

$a^2+a^2b^2+b^2+b^2c^2+c^2+a^2c^2\geq6abc$.

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LHS=$\sum a^2\ +\ \sum(ab)^2$

Applying A.M. ≥ G.M.,

$\sum a^2 ≥ 3(abc)^{\frac{2}{3}} $

$\sum (ab)^2 ≥ 3(abc)^{\frac{4}{3}} $

Taking summation,$ LHS ≥ 3((abc)^{\frac{2}{3}} + (abc)^{\frac{4}{3}}) $

But $(abc)^{\frac{2}{3}} + (abc)^{\frac{4}{3}}$ ≥ 2 abc (applying A.M. ≥ G.M.,)

LHS ≥ 3(2abc)

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