prove this inequality with five variables Let $$\begin{cases}S=a^2+b^2+c^2+d^2+e^2\\
ab+bc+cd+de+ea=T_{1}\\
ac+ce+eb+bd+da=T_{2}
\end{cases}$$
Find the range $A$,such any postive real numbers have
$$S\ge AT_{1}+(1-A)T_{2}$$
This problem is creat by Wang yong xi .since I'm using the AM-GM  inequality.
$$S=a^2+b^2+c^2+d^2+e^2\ge ab+bc+cd+de+ea=T_{1}$$
and
$$S=a^2+b^2+c^2+d^2+e^2\ge T_{2}$$
so $$S\ge \max{(T_{1},T_{2})}\ge AT_{1}+(1-A)T_{2}$$
so $A\in [0,1]?$
It is said that this is not the correct answer.
 A: Indeed, the condition is satisfied for $-0.62 \simeq \frac{1-5^{1/2} }{2}  < A< \frac43$ as will be shown below. Two examples are given.
We need to establish $S - AT_{1}-(1-A)T_{2} \ge 0 $ and we have that
$$
2[S - AT_{1}-(1-A)T_{2}] = A \sum_{cyc} (a-b)^2 + (1-A) \sum_{cyc} (a-c)^2
$$
where $\sum_{cyc}$ says that the argument is to be taken over all 5 cyclic shifts.
So as long as $0 \le A \le 1$ this is always nonnegative.
Case 1: Now consider $A<0$ and write $B = -A > 0$. Then we need to establish
$$
 \sum_{cyc} (a-c)^2 \ge \frac{B}{1+B} \sum_{cyc} (a-b)^2
$$
Case 2: Consider $A>1$ and write $C = A-1 > 0$. Then we need to establish
$$
 \sum_{cyc} (a-c)^2 \le \frac{1+C}{C} \sum_{cyc} (a-b)^2
$$
It remains to be discussed whether and for which $B,C$ these cases hold true.
Let $a-b = x$, $b-c = y$, etc. with variables $x,y,z,w,v$. Then $\sum_{cyc} x =0$ and in case 1,
$$
 \sum_{cyc} (x+y)^2 - \frac{B}{1+B} \sum_{cyc} x^2 \ge 0
$$
Now this is an expression which is unrestricted in the variables  $x,y,z,w,v$, which means we can apply calculus under the extra condition (enforced with Lagrangian $\lambda$) $\sum_{cyc} x =0$ . So let $q = \frac{B}{1+B}$ and consider 
$$
f(x,y,z,w,v) =  \sum_{cyc} (x+y)^2 - q \sum_{cyc} x^2 + \lambda \sum_{cyc} x
$$
The function is quadratic, so if we can establish global convexity, a local minimum will also be a global minimum. Global convexity is established if the Hessian is positive definite everywhere. We easily compute the Hessian
$$
H(f) = 2 \left(
\begin{matrix}
 -q + 2&     1&     0&     0&     1\\
     1& -q + 2&     1&     0&     0\\
     0&     1& -q + 2&     1&     0\\
     0&     0&     1& -q + 2&     1\\
     1&     0&     0&     1& -q + 2\\
\end{matrix}
\right)
$$
For global convexity, we need all eigenvalues of $H$ be positive. (Note that positivity of the determinant of $H$ is necessary but not sufficient.) The eigenvalues are $$\lambda_1 = 
           8 - 2q \\
 \lambda_2 = 5^{1/2} - 2q + 3\\
\lambda_3 =
 3 - 5^{1/2} - 2q
$$
where $\lambda_2$ and $\lambda_3$ are double. We obtain $q < 4$ and $q < \frac{5^{1/2} + 3}{2} \simeq 2.61$ and $q < \frac{-5^{1/2} +3}{2} \simeq 0.38$ e.g. in total  $q = \frac{B}{1+B} < \frac{-5^{1/2} +3}{2} $ or $B < \frac{q}{1-q} = \frac{5^{1/2} -1}{2} \simeq 0.62$
Case 2 can be  treated in the very same fashion. Setting $q = \frac{1+C}{C} $ we need that $
f(x,y,z,w,v) \le 0$, hence we need to establish concavity which leads to requiring that all eigenvalues of $H$ are negative, which is satisfied with $q = \frac{1+C}{C} > 4 $ or  $C< \frac13$. 
Combining the two cases gives $-0.62 \simeq \frac{1-5^{1/2} }{2}  < A< \frac43$ as the solution.
For illustration, let's consider two special settings.
1) Let $(a,b,c,d,e) = (1,1,2,2,2)$. Then $
 A \sum_{cyc} (a-b)^2 + (1-A) \sum_{cyc} (a-c)^2 =  2 A  + 4 (1-A) = 4 - 2A \ge 0
$
which is true for $A<2$, and indeed $A< \frac43$ is the sharper condition.
2) Let $(a,b,c,d,e) = (1,2,1,2,1)$. Then $
 A \sum_{cyc} (a-b)^2 + (1-A) \sum_{cyc} (a-c)^2 =  4 A  + 2 (1-A) = 2 + 2A \ge 0
$
which is true for $-1<A$, and indeed $-0.62 \simeq \frac{1-5^{1/2} }{2} < A$ is the sharper condition.
