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.
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.