This problem seems so simple, but I cannot prove it:

\(a,b,c>0\).Prove that:

\((a^2+2)(b^2+2)(c^2+2)\geq 9(ab+bc+ca)\)

I tried all sorts of inequalities, but they didn't work.I reduced it to proving

\((x+2)(y+2)(z+2)\geq9(x+y+z)\), but I cannot prove that neither.

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TopNewestHi 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 \(a=\sqrt{2}tanA\), \(b=\sqrt{2}tanB\),\(c=\sqrt{2}tanC\). Now, by using \(1+tan^{2}x=\frac{1}{cos^2x}\), We get \(\frac{4}{9}≥cosAcosBcosC(cosAsinBsinC+sinAcosBsinC+sinAsinBcosC)\), Which is same as proving \(\frac{4}{9}≥cosAcosBcosC(cosAcosBcosC-cos(A+B+C))\). 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 \(cosAcosBcosC≤(\frac{cosA+cosB+cosC}{3})^3≤cos^3y\). Thus, Finally, we are left to show that \(\frac{4}{9}≥cos^3y(cos^3y-cos(3y))\). Now, by very easy simplification , we need to show \(\frac{4}{27}≥cos^4y(1-cos^2y)\), Which follows from simple AM-GM As \((\frac{cos^2y}{2}.\frac{cos^2y}{2}.(1-cos^2y))^{\frac{1}{3}}≤\frac{1}{3}(\frac{cos^2y}{2}+\frac{cos^2y}{2}+(1-cos^2y))=\frac{1}{3}\). And the equality holds here at \(a=b=c=1\) – Dinesh Chavan · 3 years, 3 months ago

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– Bogdan Simeonov · 3 years, 3 months ago

Thanks for the quick response!Log in to reply

– Dinesh Chavan · 3 years, 3 months ago

Ok, Hope you might have cleared ur doubtsLog in to reply

– Bogdan Simeonov · 3 years, 3 months ago

Why are your comments down voted?Also, at the end of the discussion someone mentioned a pretty strong result and didn't prove it.Any ideas?Log in to reply

– Dinesh Chavan · 3 years, 3 months ago

Wait, I will try now,, For the downvotes, I know who did it, Nevermind :)Log in to reply

– Bogdan Simeonov · 3 years, 3 months ago

Ok :D .I'll try it also.Log in to reply