a,b,c >0 iar a·b·c=1 unde a,b,c∈R ; => 1+a+ab=1 +1 /bc +1 /c=bc+c +1 /bc =>a /1+a+ab=1 /bc+c+1 iar prin analogie gasim b /1+b+bc=1 /ca+a+1 si c /1+c+ca=1 /ab+b+1 <=> 1 /bc+c+1 +1 /ca+a+1 +1 /ab+b+1 =1 <=> 1/3(a·b·c /3)=1 ,ceea ce trebuia demonstrat .