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?

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