This problem seems so simple, but I cannot prove it:
a,b,c>0.Prove that:
(a2+2)(b2+2)(c2+2)≥9(ab+bc+ca)
I tried all sorts of inequalities, but they didn't work.I reduced it to proving
(x+2)(y+2)(z+2)≥9(x+y+z), but I cannot prove that neither.
#Algebra
#HelpMe!
#Inequality
#Pleasehelp
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Hi Bogdan, This question is from APMO exam. You can find the solutions here But, I can add a another solution with some trignometry here. We choose A,B,C such that a=2tanA, b=2tanB,c=2tanC. Now, by using 1+tan2x=cos2x1, We get 94≥cosAcosBcosC(cosAsinBsinC+sinAcosBsinC+sinAsinBcosC), Which is same as proving 94≥cosAcosBcosC(cosAcosBcosC−cos(A+B+C)). Now, we have some "BAD LOOKING" Trignometric terms. So lets find a way to cancel them out. OK, then letting y=A+B+C. Now bu Simple AM-GM, We Get cosAcosBcosC≤(3cosA+cosB+cosC)3≤cos3y. Thus, Finally, we are left to show that 94≥cos3y(cos3y−cos(3y)). Now, by very easy simplification , we need to show 274≥cos4y(1−cos2y), Which follows from simple AM-GM As (2cos2y.2cos2y.(1−cos2y))31≤31(2cos2y+2cos2y+(1−cos2y))=31. And the equality holds here at a=b=c=1
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Thanks for the quick response!
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Ok, Hope you might have cleared ur doubts
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