Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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 , $$\sum \sin A≤3\sin\left(\frac{A+B+C}{3}\right)=3\sin\frac{\pi}{6}=\frac{3}{2}$$

share|cite|improve this answer

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, $$a+b+c=3/2+x+y+z\le 3/2$$

share|cite|improve this answer

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 $$\begin{split}(\spadesuit)=& (\cos^2(A)+\sin^2(A)+\cos^2(B)+\sin^2(B)+1)\\-&2(\cos(A)+\cos(B)-\cos(A)\cos(B)+\sin(A)\sin(B))\\=&(\sin(A)-\sin(B))^2+(\cos(A)+\cos(B)-1)^2\geq 0.\end{split}$$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.