INAMO 2022/8 wrote: Determine the smallest positive real $K$ such that the inequality \[ K + \frac{a + b + c}{3} \ge (K + 1) \sqrt{\frac{a^2 + b^2 + c^2}{3}} \]holds for any real numbers $0 \le a,b,c \le 1$. Easily the best problem on the test, and I glad this came out on the test because this is easily the best problem on the A shortlist. We claim that $\boxed{K = \frac{1}{3} \sqrt{6}}$ works. Note that for $(a,b,c) = (1,1,0)$, we have \[ K + \frac{2}{3} \ge (K + 1) \sqrt{\frac{2}{3}} \implies K \ge \frac{1}{3} \sqrt{6} \]It suffices to prove that this choice of $K$ works. Claim. For any $0 \le a,b,c \le 1$, we have \[ \sqrt{a^2 + b^2 + c^2} \le \frac{a + b + c + \sqrt{6}}{\sqrt{2} + \sqrt{3}} \]Equality holds if and only if $(a,b,c) \in \{ (1,1,0), (1,0,1), (0,1,1), (1,1,1) \}$. Proof. Fix $a + b + c = k$. WLOG $a \ge b \ge c$. Consider the objective function $f(a,b,c) = a^2 + b^2 + c^2$. Note that \[ f(a + \varepsilon, b - \varepsilon, c) = (a + \varepsilon)^2 + (b - \varepsilon)^2 + c^2 \ge a^2 + b^2 + c^2 = f(a,b,c) \]and \[ f(a, b + \varepsilon, c - \varepsilon) = a^2 + (b + \varepsilon)^2 + (c - \varepsilon)^2 \ge a^2 + b^2 + c^2 = f(a,b,c) \]Now, consider three cases: If $0 \le k \le 1$, then by the above argument, the maximum of the objective function is achieved when $f(k,0,0)$, and we could easily check that \[ f(k,0,0) = k \le \frac{k + \sqrt{6}}{1 + \sqrt{6}} < \frac{k + \sqrt{6}}{\sqrt{2} + \sqrt{3}} \]for all $k \in [0,1]$. If $1 \le k \le 2$, then by the above argument, the maximum of the objective function is achieved when $f(1,k-1,0)$. Thus, it suffices to prove that \[ f(1,k-1,0) = \sqrt{1 + (k - 1)^2} \le \frac{k + \sqrt{6}}{\sqrt{2} + \sqrt{3}} \]However, this is true since \[ \sqrt{1 + (k - 1)^2} \stackrel{k - 1 \ge 1}{\le} (k - 1)\sqrt{2} \le \frac{k + \sqrt{6}}{\sqrt{2} + \sqrt{3}} \]and equality holds if and only if $k = 2$. If $2 \le k \le 3$, then by the above argument, the maximum of the objective function is achieved when $f(1,1,k-2)$. Therefore, it suffices to prove that \[ \sqrt{2 + (k - 2)^2} \le \frac{k + \sqrt{6}}{\sqrt{2} + \sqrt{3}} \]which is equivalent to $-2(2 + \sqrt{6})(k - 2)(k - 3) \le 0$, which is obviously true. Equality holds when $k = 2, 3$. From our claim, we have \begin{align*} \left( \frac{1}{3} \sqrt{6} + 1 \right) \sqrt{\frac{a^2 + b^2 + c^2}{3}} &= \frac{\sqrt{2} + \sqrt{3}}{3} \sqrt{a^2 + b^2 + c^2} \\ &\stackrel{\text{Claim}}{\le} \frac{a + b + c + \sqrt{6}}{3} = \frac{a + b + c}{3} + \frac{1}{3} \sqrt{6} \end{align*}and thus $K = \frac{1}{3} \sqrt{6}$ works as intended.

IndoMathXdZ wrote: INAMO 2022/8 wrote: Determine the smallest positive real $K$ such that the inequality \[ K + \frac{a + b + c}{3} \ge (K + 1) \sqrt{\frac{a^2 + b^2 + c^2}{3}} \]holds for any real numbers $0 \le a,b,c \le 1$. Easily the best problem on the test, and I glad this came out on the test because this is easily the best problem on the A shortlist. We claim that $\boxed{K = \frac{1}{3} \sqrt{6}}$ works. Note that for $(a,b,c) = (1,1,0)$, we have \[ K + \frac{2}{3} \ge (K + 1) \sqrt{\frac{2}{3}} \implies K \ge \frac{1}{3} \sqrt{6} \]It suffices to prove that this choice of $K$ works. Claim. For any $0 \le a,b,c \le 1$, we have \[ \sqrt{a^2 + b^2 + c^2} \le \frac{a + b + c + \sqrt{6}}{\sqrt{2} + \sqrt{3}} \]Equality holds if and only if $(a,b,c) \in \{ (1,1,0), (1,0,1), (0,1,1), (1,1,1) \}$. Proof. Fix $a + b + c = k$. WLOG $a \ge b \ge c$. Consider the objective function $f(a,b,c) = a^2 + b^2 + c^2$. Note that \[ f(a + \varepsilon, b - \varepsilon, c) = (a + \varepsilon)^2 + (b - \varepsilon)^2 + c^2 \ge a^2 + b^2 + c^2 = f(a,b,c) \]and \[ f(a, b + \varepsilon, c - \varepsilon) = a^2 + (b + \varepsilon)^2 + (c - \varepsilon)^2 \ge a^2 + b^2 + c^2 = f(a,b,c) \]Now, consider three cases: If $0 \le k \le 1$, then by the above argument, the maximum of the objective function is achieved when $f(k,0,0)$, and we could easily check that \[ f(k,0,0) = k \le \frac{k + \sqrt{6}}{1 + \sqrt{6}} < \frac{k + \sqrt{6}}{\sqrt{2} + \sqrt{3}} \]for all $k \in [0,1]$. If $1 \le k \le 2$, then by the above argument, the maximum of the objective function is achieved when $f(1,k-1,0)$. Thus, it suffices to prove that \[ f(1,k-1,0) = \sqrt{1 + (k - 1)^2} \le \frac{k + \sqrt{6}}{\sqrt{2} + \sqrt{3}} \]However, this is true since \[ \sqrt{1 + (k - 1)^2} \stackrel{k - 1 \ge 1}{\le} (k - 1)\sqrt{2} \le \frac{k + \sqrt{6}}{\sqrt{2} + \sqrt{3}} \]and equality holds if and only if $k = 2$. If $2 \le k \le 3$, then by the above argument, the maximum of the objective function is achieved when $f(1,1,k-2)$. Therefore, it suffices to prove that \[ \sqrt{2 + (k - 2)^2} \le \frac{k + \sqrt{6}}{\sqrt{2} + \sqrt{3}} \]which is equivalent to $-2(2 + \sqrt{6})(k - 2)(k - 3) \le 0$, which is obviously true. Equality holds when $k = 2, 3$. From our claim, we have \begin{align*} \left( \frac{1}{3} \sqrt{6} + 1 \right) \sqrt{\frac{a^2 + b^2 + c^2}{3}} &= \frac{\sqrt{2} + \sqrt{3}}{3} \sqrt{a^2 + b^2 + c^2} \\ &\stackrel{\text{Claim}}{\le} \frac{a + b + c + \sqrt{6}}{3} = \frac{a + b + c}{3} + \frac{1}{3} \sqrt{6} \end{align*}and thus $K = \frac{1}{3} \sqrt{6}$ works as intended. oh god i accidentally picked the wrong $K$ value lol, which is $K=\frac{\sqrt{3}}{3}$

Here is another proof for the claim that we have $$\sqrt{a^2 + b^2 + c^2} \le \frac{a + b + c + \sqrt{6}}{\sqrt{2} + \sqrt{3}}$$for $a,b,c\in[0, 1]$. Note that $a+b+c+\sqrt 6-(\sqrt2+\sqrt3)\sqrt{a^2+b^2+c^2}=(\sqrt2-\sqrt{a^2+b^2+c^2})(\sqrt3-\sqrt{a^2+b^2+c^2})+a+b+c-a^2-b^2-c^2$. So if $a^2+b^2+c^2\ge 2$ we are done, since $a\ge a^2,b\ge b^2,c\ge c^2$ gives that $a+b+c\ge a^2+b^2+c^2$. Now suppose $a+b+c\ge a^2+b^2+c^2\ge 2$. Let $a^2+b^2+c^2=t\in[2,3]$, then $a+b+c\ge 2+\sqrt{t-2}\Leftrightarrow (a+b+c)^2-4(a+b+c)+6\ge a^2+b^2+c^2\Leftrightarrow ab+bc+ca-2(a+b+c)+3\ge 0\Leftrightarrow (a-1)(b-1)+(b-1)(c-1)+(c-1)(a-1)\ge 0$ which is true. Thus we just need to prove that $(\sqrt2-\sqrt{t})(\sqrt3-\sqrt{t})+2+\sqrt{t-2}-t \ge0\Leftrightarrow 2+\sqrt 6\ge (\sqrt2+\sqrt3)\sqrt t-\sqrt{t-2}\Leftrightarrow \sqrt 2\ge \sqrt t-(\sqrt3-\sqrt2)\sqrt{t-2}\Leftrightarrow\sqrt2+\sqrt{3t-6}\ge \sqrt{t}+\sqrt{2t-4}$, and it's easy to verify the last ineq by squaring both sides to get $2t^2-10t+12\le 0$.

GorgonMathDota wrote: Determine the smallest positive real $K$ such that the inequality \[ K + \frac{a + b + c}{3} \ge (K + 1) \sqrt{\frac{a^2 + b^2 + c^2}{3}} \]holds for any real numbers $0 \le a,b,c \le 1$. Proposed by Fajar Yuliawan, Indonesia Assume $a\geq b\geq c.$ If $a=b=c\in\{0,1\}$ then we have nothing conclusive. Let $a+b+c=3s\in(0,3)$ and $a^2 + b^2 + c^2=3s^2+6t;t\in[0,s^2].$ If $s\in\left(0,\frac13\right]$ then from $K+s\geq(K+1)s\sqrt3$ obtain $K\geq\frac1{\sqrt3}.$ If $s\in\left[\frac13,\frac23\right]$ then $\sqrt{\frac23}+s\geq\left(\sqrt{\frac23}+1\right)\sqrt s.$ Note that since $a+b+c\geq a^2 + b^2 + c^2$ then $\left(\sqrt{\frac23}+1\right)\sqrt s\geq\left(\sqrt{\frac23}+1\right)\sqrt{\frac{a^2 + b^2 + c^2}{3}}.$ Thus, $K\geq\sqrt{\frac23}.$ If $s\in\left[\frac23,1\right)$ then by Karamata $\max t=(1-s)^2.$ Thus, $$K\geq\frac{\sqrt{3s^2-4s+2}-s}{1-\sqrt{3s^2-4s+2}}.$$Note that the function $s\to\frac{\sqrt{3s^2-4s+2}-s}{1-\sqrt{3s^2-4s+2}}$ is strictly decreasing on its domain. In conclusion, $\min K=\sqrt{\frac23}.$

Well we need to find the smallest K such that $K+1\geq \frac{1-\frac{a+b+c}{3}}{1-\sqrt{\frac{a^2+b^2+c^2}{3}}}$ which is maximizing the right hand side Notice that $(2+\sqrt{6})(a+b+c-(a^2+b^2+c^2))+(1-a)(1-b)+(1-b)(1-c)+(1-c)(1-a)\geq 0$, equality holds when two of $a,b,c$ equal $1$ and the other one equals $1$ or $0$. By expanding, we'll get $(a+b+c+\sqrt{6})^2)\geq (\sqrt{2}+\sqrt{3})^2(a^2+b^2+c^2)$ which equivalent to $1+\sqrt{\frac{2}{3}}\geq \frac{1-(\frac{a+b+c}{3})}{1-\sqrt{\frac{a^2+b^2+c^2}{3}}}$ which imply that $1+K\geq 1+\sqrt{\frac{2}{3}}$, so $K\geq \sqrt{\frac{2}{3}}$. Equality holds when $2$ of $a,b,c$ equal $1$ and the other one equals $0$ or $1$.

One of my favourite oly inequalities I have a similar solution as @above

Our answer is $\boxed{\frac{\sqrt 6}{3}}$. Notice $(1,1,0)$ gives us \[K+1 \ge (K+1) \cdot \sqrt{\frac 23}.\] We claim this is sufficent for all $a,b,c \in [0,1]$. Manipulating for $K = \frac{\sqrt 6}{3}$, we get \begin{align*} &\hspace{2cm} \frac{\sqrt 6}{3} + \frac{a+b+c}{3} \ge \left(\frac{\sqrt 6}{3} + 1\right) \sqrt{\frac{a^2+b^2+c^2}{3}} \\ &\iff 6 + 2\sqrt 6 (a+b+c) + (a+b+c)^2 \ge (5+2\sqrt 6)(a^2+b^2+c^2) \\ &\iff 3 + \sqrt 6 (a+b+c) + (ab+bc+ca) \ge (2+\sqrt 6)(a^2+b^2+c^2) \end{align*} We conclude by noting \begin{align*} 3 \ge 2&a^2 + b^2 \\ a \sqrt 6 \ge a^2 \sqrt 6, \quad b \sqrt 6 &\ge b^2 \sqrt 6, \quad c \sqrt 6 \ge c^2 \sqrt 6 \\ ab \ge b^2, \quad bc &\ge c^2, \quad ca \ge c^2.~\blacksquare \end{align*}