Let a,b,c,d be positive real numbers such that abcd = 1. Show that,?
(1 + a)(1 + b)(1 + c)(1 + d) >= 16
proof:
as a is real
(1+a) = (1-sqrt(a))^2 + 2 sqrt(a) or
1+ a >= 2 sqrt(a) ( we can also show it by AM GM inequality)
similarly
(1+b) > = 2 sqrt(b)
(1+c) > = 2 sqrt(c)
(1+d) > = 2 sqrt(d)
by multiplying
(1+a)(1+b)(1+c)(1+d)>= 16 sqrt(abcd) or > 16 as abcd = 1
3 comments:
Post more often. BTW, I am trying to popularize this blog. Posted the link on Facebook! Let the visitors descend!
And click on my name to read my poetry!
*grin*
I clicked over here thanks to your sister but I'm lucky I can even count...
Paul, you landed up at the right place. My BIG BROTHER can teach you higher math. And I CAN teach you to count.
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