Question

Demonstrati ca $9a^{2} + 8b^{2} +5c ^{2} -12ab-6ac-4bc \geq 0$ , oricare ar fi numerele reale a,b,c.

Answer 1

$6( a^{2} + b^{2}-2ab)+3( a^{2}+ c^{2}-2ac)+2( b^{2}+ c^{2}-2bc)=6(a-b)^2+$$3(a-c)^2+2(b-c)^2$≥0, evident.

Answer 2

$Este~evident~ca~trebuie~scrisa~expresia~ca~suma~de~puteri.~Asta \\ \\ se~face~prin~incercari. \\ \\ Unul~dintre~patrate~il~poate~contine~pe~3a~(ca~sa~obtinem~9a^2). \\ \\ Mai~avem~8=4+4=2^2+2^2~si~5=1+4=1^2+2^2. \\ \\ E=9a^2+8b^2+5c^2-12ab-6ac-4bc= \\ \\ =9a^2+4b^2+4b^2+c^2+4c^2-12ab-6ac-4bc= \\ \\ =(9a^2+4b^2+c^2-12ab-6ac+4bc)+(4b^2-8bc+4c^2)= \\ \\ =(3a-2b-c)^2+4(b-c)^2.$

Aceasta problema s-ar rezolva in mod normal pe o ciorna cu foarte multe mazgalituri si stersaturi (nu ies din prima patratele! Trebuie studiati coeficientii cu mare atentie).