<s>[INST] Inequality with unusual constraint $a,b,c\in (0,1)$ such that $1+abc=a(bc+a)+b(ca+b)+c(ab+c)$ 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.
 [/INST] 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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