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If `x>0,` then `(x-1)^2=x^2-2x+1>=0` , which is equivalent to `x^2+1>=2x,` which itself is equivalent to `x+1/x>=2` (remember that `x>0,` which we need to justify the last step). See the link for an alternate proof. Furthermore, if `x!=1,` then equality won't hold and we actually have `x+1/x>2.`
Anyway, if `a,b>0,` and `a!=b,` we have `(a+b)/(2sqrt(ab))> 0` and `a+b!=2sqrt(ab),` so the fraction is not 1. For simplicity, let `u=(a+b)/(2sqrt(ab)),` so `(2sqrt(ab))/(a+b)=1/u` Thus, if `a,b>=0` and `a!=b,` `u+1/u=(a+b)/(2sqrt(ab))+(2sqrt(ab))/(a+b)>2.`
Now multiply by `sqrt(ab)` to get `(a+b)/2+(2ab)/(a+b)>2sqrt(ab),` which is equivalent to what we want to prove with a slight rearrangement.
Finally, note that the above proof doesn't work if either `a` or `b` is zero, since then `u` isn't defined. But if exactly one of `a` or `b` is zero, then the left side is positive and the right zero, so the inequality is also true in that case.
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