Help me prove $a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\geq6abc$
How to prove that $a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\geq6abc$
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algebra-precalculus
inequality
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0Hint: $1 + b^2 \geq 2b$, $1 + c^2 \geq 2c$,... – 2012-07-15
2 Answers
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$a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\geq6abc$
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
$a^2+a^2b^2+b^2+b^2c^2+c^2+a^2c^2\geq6abc$.
0
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)