# Simple inequality on real numbers

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. –  Isaac Solomon Nov 16 '12 at 7:36
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. :( –  John Chang Nov 16 '12 at 7:36
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. –  Isaac Solomon Nov 16 '12 at 7:39
I am doing that sir. Thank you and very sorry. –  John Chang Nov 16 '12 at 7:40
No worries. Thanks for trying to be considerate. =D –  Isaac Solomon Nov 16 '12 at 7:44

$2(a^2 + b^2) \geq (a+b)^2 + (a-b)^2$ $≥ (a+b)^2$ . 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 .