<s>[INST] Help in proving an inequality Show that $$a^4 + b^4\ge\frac{1}{8}$$ if $a+b=1.$
 [/INST] $\text {let: } k = \frac {(b-a)}{2}\\
\text {then: } b = \frac {(b+a)}{2} + \frac {(b-a)}{2} = \frac 12 + k\\
a = \frac {(b+a)}{2} - \frac {(b-a)}{2} = \frac 12 - k$ 
$a^4 + b^4\\
(\frac 12 - k)^4 + (\frac 12 + k)^2\\
[(\frac 12)^4 - 4(\frac 12)^3k + 6(\frac 12)^2k^2 -4(\frac 12)k^3 + k^4] + [(\frac 12)^4 + 4(\frac 12)^3k + 6(\frac 12)^2k^2 +4(\frac 12)k^3 + k^4]\\
(\frac 12)^4 + 6(\frac 12)^2k^2 + k^4\\
\frac 18 + 6(\frac 12)^2k^2 + k^4$
$k^2>0, k^4 >0$ for all k.
$\frac 18 + 6(\frac 12)^2k^2 + k^4 > \frac 18$
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