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# Need help with inequality!!!

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.

Note by Bogdan Simeonov
3 years, 9 months ago

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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 $$a=\sqrt{2}tanA$$, $$b=\sqrt{2}tanB$$,$$c=\sqrt{2}tanC$$. 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 $$a=b=c=1$$

- 3 years, 9 months ago

Thanks for the quick response!

- 3 years, 9 months ago

Ok, Hope you might have cleared ur doubts

- 3 years, 9 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?

- 3 years, 9 months ago

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

- 3 years, 9 months ago

Ok :D .I'll try it also.

- 3 years, 9 months ago