Aplici inegalitatea mediilor.media aritmetica Ma>media armonica
(a+b+c)/3≥3/(1/a+1/b+1/c) inmultesti ambii membrii cu (1/a+1/b+1/.c)
(1/a+1/b+1/c)(a+b+c)/3≥3 dar cum a 2-a parenteza este 1 obtii
(1/a+1/b+1/c)/3≥3=>
1/a+1/b+1/c≥9
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(a-b)²≥0=>
a²+b²≥2ab
In mod analog arati
b²+c²≥2bc
c²+a²≥2ca Se aduna cele 3 inegalitati
2a²+2b²+2c²≥2bc+2ab+2ac imparti prin 2 si obtii
a²+b²+c²≥ab+bc+ca adica
1≥ab+bc+ca
