Как совместить две агрегатные функции в одном запросе

SELECT fio, COUNT(inq_id) AS quantity, sys_name
FROM ((employees JOIN systems ON tabnum=admin)JOIN users ON systems.sys
_id=users.sys_id) JOIN inquiries ON users.user_id=inquiries.user_id
GROUP BY sys_name, fio

SELECT fio, quantity, sys_name
FROM
(SELECT fio, sys_name, quantity, MIN(quantity) OVER (partition BY fio) min_q
FROM (здесь исходный запрос)AS a) AS b
WHERE quantity = min_q

WITH
admins AS
 (SELECT u.user_id, e.fio, s.sys_name
 FROM systems s, employees e, users u WHERE s.admin = e.tabnum AND e.tabnum = u.tabnum),
qstat AS
 (SELECT adm.fio, adm.sys_name, COUNT(q.inq_id) AS qcnt
 FROM admins adm, inquiries q WHERE adm.user_id = q.user_id
 GROUP BY adm.fio, adm.sys_name)
SELECT fio, sys_name, q1.qcnt
FROM qstat q1
WHERE qcnt = (SELECT MIN(qcnt) FROM qstat q2 WHERE q2.admin=q1.admin)
ORDER BY fio;

WITH
admins AS
 (SELECT DISTINCT s.sys_id, e.fio, s.sys_name
 FROM systems s, employees e, users u WHERE s.admin = e.tabnum AND s.sys_id = u.sys_id),
qstat AS
 (SELECT adm.fio, adm.sys_name, COUNT(q.inq_id) AS qcnt
 FROM admins adm, inquiries q, users u WHERE adm.sys_id = u.sys_id AND u.user_id=q.user_id
 GROUP BY adm.fio, adm.sys_name)
SELECT fio
FROM qstat
WHERE qcnt = (SELECT MIN(qcnt) FROM qstat)
GROUP BY  fio
ORDER BY fio;

SELECT
  s.sys_name,
  COUNT(q.inq_id)
FROM
  systems s,
  users u,
  inquiries q
WHERE
  s.sys_id = u.sys_id AND u.user_id = q.user_id
GROUP BY
  s.sys_name
HAVING COUNT(q.inq_id) = 0;