Indonesia TST IMO 2022, Fajar Yuliawan wrote: Let $a$ and $b$ be two positive reals such that the inequality \[ ax^3 + by^2 \geq xy - 1 \]holds for any positive reals $x, y \geq 1$. Determine the smallest possible value of $a^2 + b$. The answer is $\boxed{\frac{2}{3\sqrt{3}}}$, which is achieved by $(a,b) = \left( \frac{1}{\sqrt[4]{108}}, \frac{1}{2\sqrt{3}} \right)$. To show that this holds, note that we have $a^2 b^3 = \frac{1}{432}$. Therefore, for any $x, y \ge 1$, we have \begin{align*} ax^3 + by^2 &= 2 \cdot \frac{ax^3}{2} + 3 \cdot \frac{by^2}{3} + 1 - 1\\ &\ge 6 \sqrt[6]{\frac{a^2b^3 x^6 y^6}{2^2 3^3}} - 1 \\ &\ge 6xy \sqrt[6]{\frac{a^2 b^3}{2^2 3^3}} - 1 \\ &= xy - 1 \end{align*}It suffices to show that $\frac{2}{3\sqrt{3}}$ is indeed the minimum. Before that, call $(a,b)$ cool if $(a,b)$ satisfies the statement given above. We will present two solutions, where the first solution is more natural, by investigating the behavior of the polynomial, without knowing what the answer might be and the second solution is heavily dependent with the equality case in mind. Solution 1. (Analyze polynomial) Let us write $P(x,y) = ax^3 + by^2 - xy + 1$. We know that $P(x,y) \ge 0$ for all $x,y \ge 1$. The main idea is to investigate the behavior of the multivariable polynomial $P(x,y)$ as a quadratic polynomial in $y$. Claim 01. Fix $x = \ell \ge 1$. We have $Q_{\ell}(y) = P(\ell,y) \ge 0$ for all $y \in \mathbb{R} \setminus (0,1)$. Proof. Indeed, note that if $y \le 0$, we have \[ P(\ell,y) = a\ell^3 + by^2 - \ell y + 1 \ge a\ell ^3 + by^2 + 1 > 0 \]and if $y \ge 1$, we have $P(\ell,y) \ge 0$ from the condition of the problem. Now, we consider two cases: Suppose that there exists $\ell$ such that $Q_{\ell}(y)$ has two distinct real roots, which should be on interval $(0,1]$ by Claim 01. This implies \[ \frac{\ell}{2b} < \frac{\ell + \sqrt{\ell^2 - 4b(a \ell^3 + 1)}}{2b} \le 1 \]This means that $1 \le \ell < 2b \implies b > \frac{1}{2}$. Therefore, we must have $a^2 + b > \frac{1}{2} > \frac{2}{3\sqrt{3}}$. Else, $Q_{\ell}(y) \ge 0$ for all $y \in \mathbb{R}$ and $\ell \ge 1$.This implies that $Q_{\ell}(y)$ has at most one real root, i.e. $Q_{\ell}(y)$ has non-positive discriminant. We will then get \[ 4b(a x^3 + 1) \ge x^2, \ \forall x \ge 1 \]Now, call $(a,b)$ nice if $4b(ax^3 + 1) \ge x^2$ for all $x \ge 1$. We will prove that nice implies cool. Indeed, if $(a,b)$ is nice, then \[ ax^3 + by^2 = (ax^3 + 1) + by^2 - 1 \ge \frac{x^2}{4b} + by^2 - 1 \ge 2 \sqrt{\frac{x^2}{4b} \cdot by^2} - 1 = xy - 1 \]which implies that $(a,b)$ is cool. It suffices to minimize $a^2 + b$ if $(a,b)$ is nice. Let us denote $f(x) = 4abx^3 - x^2 + 4b$. We could see that $(a,b)$ nice is essentially saying that $\min_{x \ge 1} f(x) \ge 0$. Now, note that $f(x) = 4ab x^3 - x^2 + 4b$ has two critical points ($f'(x) = 0$), i.e. when $x = 0$ and $x = \frac{1}{6ab}$. Suppose that $\frac{1}{6ab} < 1$, then we have \[ a^2 + b > a^2 + \frac{1}{6a} = a^2 + 2 \cdot \frac{1}{12a} \ge 3 \sqrt[3]{a^2 \cdot \frac{1}{144a^2}} = \sqrt[3]{\frac{3}{16}} > \frac{2}{3\sqrt{3}} \] Else, $\frac{1}{6ab} \ge 1$, which means $\min_{x \ge 1} f(x) = f \left( \frac{1}{6ab} \right)$. Therefore, we have \[ f \left( \frac{1}{6ab} \right) \ge 0 \implies a^2 b^3 \ge \frac{1}{432} \]This gives us \[ a^2 + b = a^2 + 3 \cdot \frac{b}{3} \ge 4 \sqrt[4]{\frac{a^2 b^3}{27}} \ge \frac{2}{3\sqrt{3}} \]Equality holds if and only if $a^2 = \frac{b}{3}$ and $a^2 b^3 = \frac{1}{432}$. Solving which gives us $b = \frac{1}{2\sqrt{3}}$ and $a = \frac{1}{\sqrt[4]{108}}$, which we have shown to work at the beginning of the solution. Solution 2. (Tricky Manipulation) We consider two possible cases: Suppose $\max(a,b) \ge 1$, then we have $a^2 + b > 1 > \frac{2}{3\sqrt{3}}$. Else, we must have $a, b < 1$. Take $x^3 = \frac{2}{a}$ and $y^2 = \frac{3}{b}$. Clearly, $x,y > 1$. Note that \[ 6^6 = (ax^3 + by^2 + 1)^6 \ge (xy)^6 = \frac{2^2 \cdot 3^3}{a^2 \cdot b^3} \]This implies $a^2 b^3 \ge \frac{1}{432}$. Therefore, we have \[ a^2 + b = a^2 + 3 \cdot \frac{b}{3} \ge 4 \sqrt[4]{\frac{a^2 b^3}{27}} \ge \frac{2}{3\sqrt{3}} \]

Here's an elementary solution using calculus. Define $f(x,y)=ax^3+by^2-xy+1$. Taking partial derivatives, this is stationary at $3ax^2-y=0$ and $2by-x=0$, that is at $f(0,0)=1$ and $f\left(\tfrac{1}{6ab},\tfrac{1}{12ab^2}\right)=1-\tfrac{1}{432a^2b^3}$. The latter is clearly the minimum, and since it's required that this is $\ge 0$, we must have $a^2b^3\ge\tfrac{1}{432}\quad(\star)$. To minimise $a^2+b$, we want equality in $(\star)$, so $a^2=\tfrac{1}{b^3}$ and our job is now to minimise $g(b)=\tfrac{1}{432b^3}+b$. Setting $g^\prime(b)=0=\tfrac{-1}{144b^4}+1$, we find $b=\tfrac{1}{2\sqrt{3}}$, so $a=\tfrac{1}{\sqrt[4]{108}}$ and $a^2+b=\boxed{\tfrac{2}{3\sqrt{3}}}$.

IndoMathXdZ wrote: Indonesia TST IMO 2022, Fajar Yuliawan wrote: Let $a$ and $b$ be two positive reals such that the inequality \[ ax^3 + by^2 \geq xy - 1 \]holds for any positive reals $x, y \geq 1$. Determine the smallest possible value of $a^2 + b$. The answer is $\boxed{\frac{2}{3\sqrt{3}}}$, which is achieved by $(a,b) = \left( \frac{1}{\sqrt[4]{108}}, \frac{1}{2\sqrt{3}} \right)$. To show that this holds, note that we have $a^2 b^3 = \frac{1}{432}$. Therefore, for any $x, y \ge 1$, we have \begin{align*} ax^3 + by^2 &= 2 \cdot \frac{ax^3}{2} + 3 \cdot \frac{by^2}{3} + 1 - 1\\ &\ge 6 \sqrt[6]{\frac{a^2b^3 x^6 y^6}{2^2 3^3}} - 1 \\ &\ge 6xy \sqrt[6]{\frac{a^2 b^3}{2^2 3^3}} - 1 \\ &= xy - 1 \end{align*}It suffices to show that $\frac{2}{3\sqrt{3}}$ is indeed the minimum. Before that, call $(a,b)$ cool if $(a,b)$ satisfies the statement given above. We will present two solutions, where the first solution is more natural, by investigating the behavior of the polynomial, without knowing what the answer might be and the second solution is heavily dependent with the equality case in mind. Solution 1. (Analyze polynomial) Let us write $P(x,y) = ax^3 + by^2 - xy + 1$. We know that $P(x,y) \ge 0$ for all $x,y \ge 1$. The main idea is to investigate the behavior of the multivariable polynomial $P(x,y)$ as a quadratic polynomial in $y$. Claim 01. Fix $x = \ell \ge 1$. We have $Q_{\ell}(y) = P(\ell,y) \ge 0$ for all $y \in \mathbb{R} \setminus (0,1)$. Proof. Indeed, note that if $y \le 0$, we have \[ P(\ell,y) = a\ell^3 + by^2 - \ell y + 1 \ge a\ell ^3 + by^2 + 1 > 0 \]and if $y \ge 1$, we have $P(\ell,y) \ge 0$ from the condition of the problem. Now, we consider two cases: Suppose that there exists $\ell$ such that $Q_{\ell}(y)$ has two distinct real roots, which should be on interval $(0,1]$ by Claim 01. This implies \[ \frac{\ell}{2b} < \frac{\ell + \sqrt{\ell^2 - 4b(a \ell^3 + 1)}}{2b} \le 1 \]This means that $1 \le \ell < 2b \implies b > \frac{1}{2}$. Therefore, we must have $a^2 + b > \frac{1}{2} > \frac{2}{3\sqrt{3}}$. Else, $Q_{\ell}(y) \ge 0$ for all $y \in \mathbb{R}$ and $\ell \ge 1$.This implies that $Q_{\ell}(y)$ has at most one real root, i.e. $Q_{\ell}(y)$ has non-positive discriminant. We will then get \[ 4b(a x^3 + 1) \ge x^2, \ \forall x \ge 1 \]Now, call $(a,b)$ nice if $4b(ax^3 + 1) \ge x^2$ for all $x \ge 1$. We will prove that nice implies cool. Indeed, if $(a,b)$ is nice, then \[ ax^3 + by^2 = (ax^3 + 1) + by^2 - 1 \ge \frac{x^2}{4b} + by^2 - 1 \ge 2 \sqrt{\frac{x^2}{4b} \cdot by^2} - 1 = xy - 1 \]which implies that $(a,b)$ is cool. It suffices to minimize $a^2 + b$ if $(a,b)$ is nice. Let us denote $f(x) = 4abx^3 - x^2 + 4b$. We could see that $(a,b)$ nice is essentially saying that $\min_{x \ge 1} f(x) \ge 0$. Now, note that $f(x) = 4ab x^3 - x^2 + 4b$ has two critical points ($f'(x) = 0$), i.e. when $x = 0$ and $x = \frac{1}{6ab}$. Suppose that $\frac{1}{6ab} < 1$, then we have \[ a^2 + b > a^2 + \frac{1}{6a} = a^2 + 2 \cdot \frac{1}{12a} \ge 3 \sqrt[3]{a^2 \cdot \frac{1}{144a^2}} = \sqrt[3]{\frac{3}{16}} > \frac{2}{3\sqrt{3}} \] Else, $\frac{1}{6ab} \ge 1$, which means $\min_{x \ge 1} f(x) = f \left( \frac{1}{6ab} \right)$. Therefore, we have \[ f \left( \frac{1}{6ab} \right) \ge 0 \implies a^2 b^3 \ge \frac{1}{432} \]This gives us \[ a^2 + b = a^2 + 3 \cdot \frac{b}{3} \ge 4 \sqrt[4]{\frac{a^2 b^3}{27}} \ge \frac{2}{3\sqrt{3}} \]Equality holds if and only if $a^2 = \frac{b}{3}$ and $a^2 b^3 = \frac{1}{432}$. Solving which gives us $b = \frac{1}{2\sqrt{3}}$ and $a = \frac{1}{\sqrt[4]{108}}$, which we have shown to work at the beginning of the solution. Solution 2. (Tricky Manipulation) We consider two possible cases: Suppose $\max(a,b) \ge 1$, then we have $a^2 + b > 1 > \frac{2}{3\sqrt{3}}$. Else, we must have $a, b < 1$. Take $x^3 = \frac{2}{a}$ and $y^2 = \frac{3}{b}$. Clearly, $x,y > 1$. Note that \[ 6^6 = (ax^3 + by^2 + 1)^6 \ge (xy)^6 = \frac{2^2 \cdot 3^3}{a^2 \cdot b^3} \]This implies $a^2 b^3 \ge \frac{1}{432}$. Therefore, we have \[ a^2 + b = a^2 + 3 \cdot \frac{b}{3} \ge 4 \sqrt[4]{\frac{a^2 b^3}{27}} \ge \frac{2}{3\sqrt{3}} \]

Nice! Your second solution is the same as the official solution.