Solution of equation x⋅20141x+1x⋅2014x2=2014
MyTry:: Clearly Here x>0, Now Using A.M≥G.M
So Here x⋅20141x>0 and 1x⋅2014x>0 .
So (x⋅20141x+1x⋅2014x2)≥(x⋅20141x⋅1x⋅2014x)12=√2014(x+1x)=2014
And equality Hold , When x=1x⇒x=1
Can we solve it without Using A.M≥G.M
If Yes, Then please explain me.
Thanks
Answer
Applying AM-GM probably gives the most elegant solution (and indeed the presentation is quite suggestive of AM-GM), but this can also be solved by using an idea behind AM-GM. Noting first that x≥0, we have:
x⋅20141x+1x⋅2014x2=2014⟹x20141x−1+1x2014x−1=2⟹(√x20141x−1−√1x2014x−1)2+2√20141x+x−2=2⟹(√x20141x−1−√1x2014x−1)2+2√2014(√1x−√x)2=2⟹2√2014(√1x−√x)2≤2⟹2014(√1x−√x)2≤1⟹(√1x−√x)2≤0⟹(√1x−√x)2=0⟹√1x−√x=0⟹1x=x⟹x=1
where we have used the fact that x≥0 several times.
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