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Let $a, b > 0$ such that $a + b ≥ 1$. Show that $a^4 + b^4 ≥ \frac18$.

What is the best possible approach on this problem?

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    Mr. Chang, it appears you have the pernicious habit of not accepting any answers. This provides a disincentive for many of us to try and help you.2012-11-16
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    That's what I'm doing now sir. Do I just click the up button on their answer? I'm sorry sir. I didn't know that feature. :(2012-11-16
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    To the left of the answer, where the up and down arrows are, as well as the score, there should be a small green arrow. Click on this green arrow beside the solution you wish to accept.2012-11-16
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    I am doing that sir. Thank you and very sorry.2012-11-16
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    No worries. Thanks for trying to be considerate. =D2012-11-16

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$2(a^2 + b^2) \geq (a+b)^2 + (a-b)^2$ $ ≥ (a+b)^2 $ . Similarly we have , $a^4+b^4≥ \frac {(\space a^2\space+\space b^2\space)^2}2$

Hence we get $$a^4+b^4≥ \frac {(\space a\space+\space b\space)^4}8$$ ; the rest follows easily .