1requesting users to post more inequalities...
6 Answers
49the question::
if\; a\geq b>0 \; \; \; \; \Rightarrow a+\frac{a}{b(a-b)}\geq 4
62Good question that I wouldnt have thought had prophet sir not posted the soln the other day ;)
this one is (a-b)+b+\frac{(a-b)+b}{b(a-b)}=(a-b)+b+\frac1b+\frac1{a-b}
Use AM GM
or else use t+1/t>=2
for a-b and b seperately...
(Edit: latexified).
341Nishant, nice solution
I had a^2 = (a-b+b)^2 \ge 4b(a-b) \Rightarrow \frac{a}{b(a-b)} \ge \frac{4}{a}
\Rightarrow a + \frac{a}{b(a-b)} \ge a + \frac{4}{a} \ge 4
1i did the same
put a-b=k ≥0 →a=b+k
so problem reduces to (b+k)+(b+k)/bk=(b+k)(1+1/bk)≥4