Prove that (a−2)(b−2)(c−2)≤1.
My solution:
Note that
(a−2)(b−2)(c−2)
=abc(a−2a)(b−2b)(c−2c)
=abc(1−2a)(1−2b)(1−2c)
=abc(1a+1b+1c−2a)(1a+1b+1c−2b)(1a+1b+1c−2c)
=abc(1b+1c−1a)(1a+1c−1b)(1a+1b−1c)
We now use the famous identity that says for all real and positive x,y and z, we have
xyz≥(x+y−z)(x+z−y)(y+z−x)
In our case, we have x=1a,y=1b,z=1c and so we get
1abc≥(1a+1b−1c)(1a+1c−1b)(1b+1c−1a)
i.e.
abc(1a+1b−1c)(1a+1c−1b)(1b+1c−1a)≤1
The proof is then follows.
Equality occurs when a=b=c=3.
Well done Belle. Good job! :-))
ReplyDeleteAww...thank you Michelle for your nice compliment, I appreciate it!
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