Is $L=\{a^ib^jc^k : i\leq j\leq k\}$ context-free?
A word like $a^nb^nc^n$ should work. Remember that the pumping lemma also allows you to pump down, i.e., delete the pumping segments. 

Also, each splitting $z=uvxyz$ can have its own "contradiction" to show it does not adhere to the pumping lemma. You can use different arguments, so different $i$ for each splitting. That is very natural here as $a$ has the minimal number of occurrences, and contradiction is reached by adding occurrences, whereas for $c$ deletion (choosing $i=0$) will get you into trouble.