Given positive numbers $a, b, c, x, y, z$, such that $a + x = b + y = c + z = S$, prove that $ay + bz +cx < S^2$ Given positive numbers $a, b, c, x, y, z$, such that $a + x = b + y = c + z = S$, prove that $ay + bz +cx < S^2$
One solution is:
Denote $T = S/2$. One of the triples $(a, b, c)$ and $(x, y, z)$ has the property that at least two of its members are greater than or equal to $T$ . Assume that $(a, b, c)$ is the one, and choose $\alpha = a - T, \beta = b-T$ and $\gamma = c-T$. We then have $x = T-\alpha, y = T - \beta$ and $z= T - \gamma$.
Now the required inequality is equivalent to
$$
(T+\alpha)(T-\beta)+(T+\beta)(T-\gamma)+(T+\gamma)(T-\alpha) < 4T^2
$$
After simplifying we get that what we need to prove is
$$
-(\alpha \beta + \beta \gamma + \gamma \alpha) <T^2  \ \ \ \ \ \ (1)
$$
We also know that at most one of the numbers $\alpha, \beta, \gamma$ is negative. If all are positive, there is nothing to prove. Assume that $\gamma < 0$. Now (1) can be rewritten as $-\alpha \beta -\gamma(\alpha+\beta) < T^2$. Since $-\gamma < T$ we have that $-\alpha\beta - \gamma(\alpha+\beta) < -\alpha\beta +T(\alpha+\beta)$ and the last term is less tham $T$ since $(T-\alpha)(T-\beta)>0$
Now, I'm looking for other solutions to prove it, please comment on
 A: The system is homogenous, so WLOG $S=1$. Now all numbers are from $(0,1)$ and we want to prove
$$a(1-b)+b(1-c)+c(1-a)<1$$
Substitute $a=1/(1+p)$, $b=1/(1+q)$, $c=1/(1+r)$, where $p,q,r\in(0,\infty)$. It becomes
\begin{align}\frac1{1+p}\frac q{1+q}+\frac1{1+q}\frac r{1+r}+\frac1{1+r}\frac p{1+p}&<1\\
q(1+r)+r(1+p)+p(1+q)&<(1+p)(1+q)(1+r)\\
p+q+r+pq+qr+rp&<1+p+q+r+pq+qr+rp+pqr\\
0&<1+pqr\end{align}
And we are done.

You can find more solutions here, it's basically the same question: Find the maximum value for $x+y+z-xy-yz-zx$
A: Assmue equilateral triangle $PQR$,and denoting the common length of the sides of the equilateral triangle as $k$,and 
$$QL=x,LR=a,RM=y,MP=b,PN=z,NQ=c$$
then it is clear
$$S_{LRM}+S_{MPN}+S_{NQL}<S_{PQR}$$
then we have
$$\dfrac{1}{2}ay\sin{60^{0}}+\dfrac{1}{2}bz\sin{60^{0}}+\dfrac{1}{2}cx\sin{60^{0}}
<\dfrac{1}{2}k^2\sin{60^{0}}$$
so
$$ay+bz+cx<k^2$$
A: Here's a direct solution based on the idea of finding the volume of a cube
$$ S ( ay + bz + cx ) = ay (c+z) + bz (a  + x) + cx ( b+y) \leq ( a + x)( b+y)( c + z) = S^3  $$
hence, $ ay + bz + cz \leq S^3 $.
Furthermore, strict inequality holds since $abc >0, xyz > 0$.

See here for more problems where the inequality is solved with a geometric interpretation.
