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What is the proof that for any acute triangle $ABC$,then :

$$\cos^3 (A)+ \cos^3 (B)+\cos^3 (C)+\cos(A)\cdot\cos(B)\cdot\cos(C)\ge\frac{1}{2} $$

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I want to write down a full complete solution, because I think has been very instructive for me and I want to share my thoughts with everybody.

To begin with, as Tapu said, we use AM-GM to prove that $$\cos^3 x+\frac{\cos x}{4}\geq \cos^2 x.$$ Why one would do that? Well, basically it is a matter of equality case. The inequality proposed is an equality in the case of an equilateral triangle, and in that case one has that AM-GM is in fact an equality. Well, then we get that $$\operatorname{LHS}\geq \cos^2(A)+\cos^2(B)+\cos^2(C)+2\cos(A)\cos(B)\cos(C)-\cos(A)\cos(B)\cos(C)-\frac14\left(\cos(A)+\cos(B)+\cos(C)\right)\stackrel{?}{\geq}\frac12\quad (\clubsuit)$$ And I remark that the last inequality is still a guess to be proven. But now

Theorem 1: Let $ABC$ be a triangle. Then $$\cos^2(A)+\cos^2(B)+\cos^2(C)+2\cos(A)\cos(B)\cos(C)=1.$$

Proof: $$\begin{split}&\cos(A+B)=\cos(\pi-C)=-\cos(C)\\ \Rightarrow &\cos(A)\cos(B)-\sin(A)\sin(B)=-\cos(C)\\\Rightarrow &\cos(A)\cos(B)+\cos(C)=\sin(A)\sin(B)\\\Rightarrow&(\cos(A)\cos(B)+\cos(C))^2=(1-\cos^2(A))(1-\cos^2(B)),\end{split}$$ from which the thesis follows.

Theorem 2: Let $ABC$ be a triangle. Then $$\cos(A)\cos(B)\cos(C)\leq \frac 18.$$

Proof: it is easy to see that the only interesting case is when $ABC$ is an acute triangle, the other cases being trivial. By AM-GM and Jensen inequality, since the cosine is concave on $(0,\frac\pi2)$, one gets that $$\sqrt[3]{\cos(A)\cos(B)\cos(C)}\leq \frac13(\cos(A)+\cos(B)+\cos(C))\leq \cos\left(\frac{A+B+C}{3}\right)=\frac12.$$ Raise everything to the third power to get the desired result.

Theorem 3: Let $ABC$ be an acute triangle. Then $$\cos(A)+\cos(B)+\cos(C)\leq \frac 32.$$

Again this is a simple consequence of Jensen inequality plus the concavity of the cosine on $(0,\frac\pi 2)$. Therefore $$\cos(A)+\cos(B)+\cos(C)\leq 3\cos\left(\frac{A+B+C}{3}\right)=\frac 32.$$

Summing up everything, and turning back to $(\clubsuit)$, we finally conclude that $$\operatorname{LHS}\geq 1-\frac 18-\frac 38=\frac 12.$$

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But after subtracting, i mean this $2\cos(A)\cos(B)\cos(C)-\cos(A)\cos(B)\cos(C)$ there will be a term not included in the beginning of the solution after applying AM-GM. – Frank Sep 28 '12 at 12:55
@MohammedAl-mubark wait... AM-GM brings the square terms - the terms with $1/4$. I've then rewritten the term $\prod \cos=2\prod \cos- 1\prod\cos$. Is it clear? – uforoboa Sep 28 '12 at 13:09
Yes ,very clear thanks . – Frank Sep 28 '12 at 17:29

Hints: Note that for acute triangle, cosines of all the angles are positive. So, by Arithmetic–Geometric means inequality $$\cos^3 x+\frac{\cos x}{4}\ge \cos^2 x\Leftrightarrow\cos^3 x\ge \cos^2 x-\frac{\cos x}{4},\quad\forall x=A,B,C$$

Hence we are left with the following quantity, $$\sum \cos^2x -\frac{1}{4}\sum \cos x +\Pi \cos x$$

Now there are many useful things, you have to use some of them:

  • For any triangle, $\sum \cos^2x +2\Pi \cos x=1$
  • $\sum\cos x\le \frac{3}{2}$
  • $\sum\cos^2 x\ge \frac{3}{4}$
  • $\Pi\cos x\le \frac{1}{8}$

A Remark: Equality holds iff $A=B=C$

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