Question

simple inequality

Answer 1

05 Q11?

Answer 2

You've hidden something in the question :P

Answer 3

??

Answer 4

yes, but can do it:)

Answer 5

2 marks only... :(

Answer 6

kinda looks like all a.m. and g.m. to me.

Answer 7

Nope :P

Answer 8

simple inequality lol:)

Answer 9

hint: use A.M. >= G.M.

Answer 10

i was thinking that ... can't find any sol. for three vars.

Answer 11

lol ... i don't remember proving this for three var ...having prob with that..

Answer 12

just prove that\[a+b+c-3\sqrt[3]{abc} \ge a+b-2\sqrt{ab}\]

Answer 13

AM >= GM funda. Apply.

Answer 14

yes, AM>=GM\[(c+\sqrt{ab} +\sqrt{ab})/3 \ge \sqrt[3]{c \sqrt{ab}\sqrt{ab}}\]\[c+\sqrt{ab} +\sqrt{ab} \ge 3\sqrt[3]{abc}\]\[c+2\sqrt{ab} \ge 3\sqrt[3]{abc}\]\[a+b+c-3\sqrt[3]{abc} \ge a+b-2\sqrt{ab}\]