Fie a,b>0. Atunci 1/a+1/b>=4/(a+b).
DEMONSTRATIE
1/a+1/b>=4/(a+b) <=> a+b>=4ab/(a+b) <=> (a+b)^2>=4ab <=> a^2+2ab+b^2>=4ab <=> a^2-2ab+b^2>=0 <=> (a-b)^2>=0, adevarat
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x,y,z>0.
1/x+1/y+1/z>=2(1/(x+y)+1/(y+z)+1/(z+x)) <=> 2/x+2/y+2/z>=4/(x+y)+4/(y+z)+4/(z+x) <=> (1/x+1/y)+(1/y+1/z)+(1/z+1/x)>=4/(x+y)+4/(y+z)+4/(z+x), care este adevarata pe baza inegalitatii demonstrate mai sus (1/a+1/b>=4/(a+b)).
