Inequality Proof Problems [401-450]
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P401. Autocorrelation Inequality
Let \(f \ge 0\) and suppose the support of \(f\) is contained in \([-\tfrac14,\tfrac14]\). Assume that, for all such functions \(f\), the following inequality holds:
\[\max_{|t|\le \tfrac12} (f * f)(t) \ge C_1\left(\int f\right)^2.\]Find an upper bound for \(C_1\). Note: This is essentially a “min–max” problem (minimizing the worst-case / largest value).
A401. The best currently known upper bound is
\[C_1 \le 1.50286.\]See ref. Future progress may further lower this upper bound.
P402. Autocorrelation Inequality
Let \(f \ge 0\). Define
\[C_2=\sup_{f\ge 0}\frac{\|f*f\|_2^2}{\|f*f\|_1\,\|f*f\|_\infty}.\]Find a lower bound for \(C_2\). Note: This is essentially about achieving a large value (maximization via the supremum).
A402. The best currently known lower bound is
\[C_2 \ge 0.9610.\]See ref). Future progress may further raise this lower bound.
P403. Let $ABCD$ be a cyclic quadrilateral. Determine the smallest constant $C$ such that the following inequality holds if and only if $AB \cdot BC = 2AD \cdot DC$:
\[BD^2 \leq C \cdot AC^2\]P404. Let $n$ be an integer greater than 2. Find the greatest real number $C_{\text{min}}$ and the least real number $C_{\text{max}}$ such that for any positive real numbers $x_1, x_2, \ldots, x_n$ (with $x_n = x_0$ and $x_{n+1} = x_1$), the following inequality holds:
\[C_{\text{min}} \leq \sum_{i=1}^{n} \frac{x_i}{x_{i-1} + 2(n-1)x_i + x_{i+1}} \leq C_{\text{max}}\]S404. $C_{\text{min}} = \frac{1}{2(n-1)}$, $C_{\text{max}} = \frac{1}{2}$
We will prove that $m_{n}=\frac{1}{2(n-1)}, M_{n}=\frac{1}{2}$. First, let us see that the inequality
\[\\sum_{i=1}^{n} \\frac{x_{i}}{x_{i-1}+2(n-1) x_{i}+x_{i+1}} \\geq \\frac{1}{2(n-1)}\]is trivial, because $x_{i-1}+2(n-1) x_{i}+x_{i+1} \leq 2(n-1) \cdot \sum_{k=1}^{n} x_{k}$ for all $i$. This shows that $m_{n} \geq \frac{1}{2(n-1)}$. Taking $x_{i}=x^{i}$, the expression becomes
\[\\frac{1}{x+x^{n-1}+2(n-1)}+\\frac{(n-2) x}{1+2(n-1) x+x^{2}}+\\frac{x^{n-1}}{1+2(n-1) x^{n-1}+x^{n-2}}\]and taking the limit when $x$ approaches 0, we find that $m_{n} \leq \frac{1}{2(n-1)}$ and thus $m_{n}=\frac{1}{2(n-1)}$
Now, we will prove that $M_{n} \geq \frac{1}{2}$. Of course, it suffices to prove that for any $x_{1}, x_{2}, \ldots, x_{n}>0$ we have
\[\\sum_{i=1}^{n} \\frac{x_{i}}{x_{i-1}+2(n-1) x_{i}+x_{i+1}} \\leq \\frac{1}{2}\]But it is clear that
\[\\begin{gathered}\n\\sum_{i=1}^{n} \\frac{2 x_{i}}{x_{i-1}+2(n-1) x_{i}+x_{i+1}} \\leq \\sum_{i=1}^{n} \\frac{2 x_{i}}{2 \\sqrt{x_{i-1} \\cdot x_{i+1}}+2(n-1) x_{i}}= \\\\\n=\\sum_{i=1}^{n} \\frac{1}{n-1+\\frac{\\sqrt{x_{i-1} \\cdot x_{i+1}}}{x_{i}}}\n\\end{gathered}\]Taking $\frac{\sqrt{x_{i-1} \cdot x_{i+1}}}{x_{i}}=a_{i}$, we have to prove that if $\prod_{i=1}^{n} a_{i}=1$ then $\sum_{i=1}^{n} \frac{1}{n-1+a_{i}} \leq 1$. But this has already been proved in the problem 84. Thus, $M_{n} \geq \frac{1}{2}$ and because for $x_{1}=x_{2}=\cdots=x_{n}$ we have equality, we deduce that $M_{n}=\frac{1}{2}$, which solves the problem.
Equality in the lower bound $C_{\text{min}} = \frac{1}{2(n-1)}$ is achieved in the limit as one variable tends to zero and the others are fixed, for example, by taking $x_1 = x_2 = \cdots = x_{n-1} = 1$ and $x_n \to 0$. Equality in the upper bound $C_{\text{max}} = \frac{1}{2}$ is achieved when all variables are equal, i.e., $x_1 = x_2 = \cdots = x_n$. Thus, $C_{\text{min}}$ is the minimum value and $C_{\text{max}}$ is the maximum value of the given sum.
Therefore, the answer is \(C_{\\text{min}} = \\frac{1}{2(n-1)}\)\n\(C_{\\text{max}} = \\frac{1}{2}\).
P405. Let $a, b, c$ be the lengths of the sides of a triangle. Find the constant $C$ such that the following equation holds for all $a, b, c$:
\[2(a b^{2} + b c^{2} + c a^{2}) = a^{2} b + b^{2} c + c^{2} a + C a b c\]and ensures that the triangle is equilateral.
S405. $C = 3$
We’ll show that
\[a^{2} b+b^{2} c+c^{2} a+3 a b c \\geq 2\\left(a b^{2}+b c^{2}+c a^{2}\\right)\]with equality if and only if $a=b=c$, i.e. the triangle is equilateral.\nLet us use Ravi’s substitutions, i.e. $a=x+y, b=y+z, c=z+x$. Then the given inequality becomes
\[x^{3}+y^{3}+z^{3}+x^{2} y+y^{2} z+z^{2} x \\geq 2\\left(x^{2} z+y^{2} x+z^{2} y\\right)\]Since $AM \geq GM$ we have
\[x^{3}+z^{2} x \\geq 2 x^{2} z,\\quad y^{3}+x^{2} y \\geq 2 y^{2} x,\\quad z^{3}+y^{2} z \\geq 2 z^{2} y\]After adding these inequalities we obtain
\[x^{3}+y^{3}+z^{3}+x^{2} y+y^{2} z+z^{2} x \\geq 2\\left(x^{2} z+y^{2} x+z^{2} y\\right)\]Equality holds if and only if $x = y = z$, which means $a = b = c$, i.e., the triangle is equilateral. This gives the unique value $C = 3$ for which the equation holds for all $a, b, c$ and ensures the triangle is equilateral.\n\nTherefore, the answer is $C = 3$.
Related Theorem: Let $a, b, c \in \mathbb{R}^{+}$, and let us denote
\[\\begin{aligned} QM & =\\sqrt{\\frac{a^{2}+b^{2}+c^{2}}{3}}, \\quad AM=\\frac{a+b+c}{3}, \\quad GM=\\sqrt[3]{a b c} \\text { and } \\\\HM & =\\frac{3}{\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}}\\end{aligned}\]Then
\[QM \\geq AM \\geq GM \\geq HM\]Equalities occur if and only if $a=b=c$.”
Input: 2026.01.22 22:08