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Suppose as in the title that $a,b,c$ are three real positive numbers in $(0,1)$ such that $1+abc=a(bc+a)+b(ca+b)+c(ab+c)$. Then I was asked to prove that $$a+b+c\leq \frac 32.$$ I am not very good at inequalities, so can anybody help me with this task? Especially the first condition seems very unusual to me. Thanks.

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Your equation rearranges to give $(a+b+c)^2 = 1 + 2(ab+bc+ca-abc)$, and so (since everything's positive etc) this is equivalent to proving $ab+bc+ca-abc \le \frac{5}{8}$. This might help; but I'm not very good at inequalities either, so I might be leading you in the wrong direction! –  Clive Newstead Sep 11 '12 at 14:45

3 Answers 3

up vote 2 down vote accepted

As $a,b,c \in (0,1)$, let $a=\cos A, b=\cos B,c=\cos C$, so that $0<A,B,C<\frac{\pi}{2}$

So, $\cos^2A+\cos A(2\cos B\cos C)+\cos^2B+\cos^2C-1=0$

$$\implies \cos A=\frac{-2\cos B\cos C±\sqrt{(2\cos B\cos C)^2-4\cdot 1\cdot (\cos^2B+\cos^2C-1)}}{2}$$

$$\implies \cos A=-\cos B\cos C±\sin B\sin C$$ as $(2\cos B\cos C)^2-4(\cos^2B+\cos^2C-1)=4(1-\cos^2B)(1-\cos^2C)=4\sin^2B\sin^2C$

Taking the '+' sign, $ \cos A=-\cos B\cos C+\sin B\sin C=-\cos(B+C)$ $=\cos(\pi±(B+C))$ or $A+B+C=\pi$ as $0<A,B,C<\frac{\pi}{2}$

If $f(x)=\cos x,f'(x)=-\sin x, f''(x)=-\cos x<0$ as $0<x<\frac{\pi}{2}$

So, $\cos x$ is concave function in $(0,\frac{\pi}{2})$.

So using Jensen's inequality, $$\sum \cos A≤3\cos\left(\frac{A+B+C}{3}\right)=3\cos\frac{\pi}{3}=\frac{3}{2}$$

Taking the '-' sign, $ \sin A=-\sin B\sin C-\cos B\cos C$ $=-\cos(B-C) =\cos(\pi±(B-C))$ $\implies A-B+C=\pi$ or $A+B-C=\pi$ which is impossible as $0<A,B,C<\frac{\pi}{2}$.

Alternatively, we can put $a=\sin A,$ etc., where $0<A,B,C<\frac{\pi}{2}$.

Then $\sin A = \cos(B+C)=\sin(\frac{\pi}{2}±(B+C))\implies A+B+C=\frac{\pi}{2}$, as $A-B-C=\frac{\pi}{2}$ is not allowed as $0<A,B,C<\frac{\pi}{2}$.

Again, $\sin x $ is concave function in $(0,\frac{\pi}{2})$.

So using Jensen's inequality , $$\sum \sin A≤3\sin\left(\frac{A+B+C}{3}\right)=3\sin\frac{\pi}{6}=\frac{3}{2}$$

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It is easy to see that the equality is achieved at a=b=c=1/2. Therefore, let's expand around that point, i.e. define $x,y,z$ by $a=x+1/2$, $b=y+1/2$, $c=z+1/2$. Then, the first equality becomes $$1=2abc+a^2+b^2+c^2\\ \Longleftrightarrow 0=2xyz+xy+yz+xz+x^2+y^2+z^2+\frac{3}{2}(x+y+z)$$ Next, Jensen's inequality gives you that the geometric mean is smaller than the arithmetic mean, i.e. $$(x^2y^2z^2)^{1/3}\le \frac{x^2+y^2+z^2}{3}\\ \Longleftrightarrow |xyz|\le \frac{x^2+y^2+z^2}{3}\sqrt{\frac{x^2+y^2+z^2}{3}}\le\frac{x^2+y^2+z^2}{6}$$ where the last inequality follows from the fact that $-1/2<x,y,z<1/2$. Therefore, going back to the equality above $$0\ge -\frac{x^2+y^2+z^2}{3}+xy+yz+xz+x^2+y^2+z^2+\frac{3}{2}(x+y+z)\ge\\ \ge -\frac{x^2+y^2+z^2}{2}+xy+yz+xz+x^2+y^2+z^2+\frac{3}{2}(x+y+z)=\\ =\frac{2xy+2xz+2zy+x^2+y^2+z^2}{2}+\frac{3}{2}(x+y+z)\\ \Longleftrightarrow x+y+z\le -\frac{(x+y+z)^2}{3}\le 0$$ Thus, $$a+b+c=3/2+x+y+z\le 3/2$$

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EDIT.. Probably the following answer does not adress the problem the OP asked..

The unusual condition you are pointing at it's so unusual that makes you try with something obvious I think.

To show what I mean I would like to set $$\begin{cases}a=\cos(A),\\ b=\cos(B),\\ c=\cos(C).\end{cases}$$ Why that? Well, just because $a,b,c\in (0,1)$. But then one tries something with addition formulas and what can be found is that $$\begin{split}(\clubsuit)\: \cos(A+B+C)=&\cos(A)\cos(B)\cos(C)-\sin(A)\sin(B)\cos(C)\\-&\sin(A)\cos(B)\sin(C)-\cos(A)\sin(B)\sin(C).\end{split}$$ Now, what if $A+B+C=\pi$?

Well, we have $$-\cos(C)=\cos(\pi-C)=\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B),$$ and similarly for the other angles, so that, substituting in $(\clubsuit)$, after multiplying everything by $-1$ we end up with $$\begin{split}\cos(A)\cos(B)\cos(C)+1=&\cos(A)(\cos(B)\cos(C)+\cos(A))+\\&\cos(B)(\cos(C)\cos(A)+\cos(B))+\\ &\cos(C)(\cos(A)\cos(B)+\cos(C)),\end{split}$$ which is exactly the condition given at the beginning.

Then the substitution is applicable and the problem is asking to show that if $A+B+C=\pi$, then $$\cos(A)+\cos(B)+\cos(C)\leq\frac 32.$$ We substitute $$\cos(C)=\cos(\pi-(A+B))=-\cos(A+B)=-\cos(A)\cos(B)+\sin(A)\sin(B),$$ and we find that we have to prove is $$(\spadesuit)\:3-2(\cos(A)+\cos(B)-\cos(A)\cos(B)+\sin(A)\sin(B))\geq 0.$$ This is true because $$\begin{split}(\spadesuit)=& (\cos^2(A)+\sin^2(A)+\cos^2(B)+\sin^2(B)+1)\\-&2(\cos(A)+\cos(B)-\cos(A)\cos(B)+\sin(A)\sin(B))\\=&(\sin(A)-\sin(B))^2+(\cos(A)+\cos(B)-1)^2\geq 0.\end{split}$$

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I don't think $A+B+C = \pi$ is given –  Cocopuffs Sep 11 '12 at 15:13
    
Maybe you're right.... I didn't think about that.. I am not even sure the two conditions are equivalent.. –  uforoboa Sep 11 '12 at 15:50

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