11x2+y2>0
=>(x+y)2-2xy>0
=>(x+y)2>2xy>xy
=>x+yxy>1x+y
=>1x+1y>1x+y (proved)
5 Answers
11
262take lcm and then cross multiply ,
you will get x2+xy+y2 >0 which is true for all x,y >0
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or use AM>HM
x+y2≥ 2xyx+y
or, 2x+y≤ 1/2*(1x+1y)
hence 1x+y< (1x+1y)
341but with x,y>0
\frac{1}{x}>\frac{1}{x+y}
so its pretty obvious.
Maybe you wanted to prove that:
\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}
1for the above question,
replacing x,y with 1x and 1y ,
→ we have to prove (x+y)2 ≥4xy
→ which is nothing but (x-y)^2 ≥ 0