Sketch (for just the following claim; for the rest it's a full proof). By various methods (Lagrange multipliers is the simplest and quickest), one can reach the conclusion the only triplets $(a,b,c)$ worth verifying (critical) are $\bullet$ $a=b=c=\sqrt[3]{x}$ for some $x\in [0,1]$, when $\textrm{LHS} = \dfrac {3\sqrt[3]{x}}{x+1} + \sqrt[n]{x}$; $\bullet$ $c=1$, $a=b=\sqrt{x}$ for some $x\in [0,1]$, when $\textrm{LHS} = \dfrac {2\sqrt{x}+1}{x+1} + \sqrt[n]{x}$; $\bullet$ $b=c=1$, $a=x$ for some $x\in [0,1]$, when $\textrm{LHS} = \dfrac {x+2}{x+1} + \sqrt[n]{x}$. But $2\sqrt{x}+1 \leq x+2$, since equivalent to $(\sqrt{x}-1)^2 \geq 0$, and $3\sqrt[3]{x} \leq 2\sqrt{x}+1$, since equivalent to $(\sqrt[6]{x}-1)^2(2\sqrt[6]{x} + 1) \geq 0$, thus the only one worth to be checked is $\dfrac {x+2}{x+1} + \sqrt[n]{x} \leq \dfrac {5}{2}$, equivalent to $\dfrac {1}{x+1} + \sqrt[n]{x} \leq \dfrac {3}{2}$. This expression on the $\textrm{LHS}$ is clearly increasing in $n$. Now, for $n=3$, denoting $t=\sqrt[3]{x}$, it is equivalent to $\dfrac {1}{t^3+1} + t \leq \dfrac {3}{2}$, or $2t^4 - 3t^3 + 2t - 1 \leq 0$, or again $(t-1)(2t^3-t^2-t+1)\leq 0$. But $0\leq t\leq 1$, and $2t^3-t^2-t+1 = 2t(t-1/2)^2 + (t-3/4)^2 + 7/16 > 0$, so the inequality holds true. However, for $n=4$, denoting $t=\sqrt[4]{x}$, it would be equivalent to $\dfrac {1}{t^4+1} + t \leq \dfrac {3}{2}$, or $2t^5 - 3t^4 + 2t - 1 \leq 0$, or again $(t-1)^2(2t^3+t^2-1)\leq 0$, clearly false for $t<1$ but $t$ close enough to $1$. Therefore the largest such $n$ is $\boxed{n=3}$. A nasty enough problem that it was a good thing it was not selected for the competition.

mavropnevma wrote: Sketch (for just the following claim; for the rest it's a full proof). By various methods (Lagrange multipliers is the simplest and quickest), one can reach the conclusion the only triplets $(a,b,c)$ worth verifying (critical) are $\bullet$ $a=b=c=\sqrt[3]{x}$ for some $x\in [0,1]$, when $\textrm{LHS} = \dfrac {3\sqrt[3]{x}}{x+1} + \sqrt[n]{x}$; $\bullet$ $c=1$, $a=b=\sqrt{x}$ for some $x\in [0,1]$, when $\textrm{LHS} = \dfrac {2\sqrt{x}+1}{x+1} + \sqrt[n]{x}$; $\bullet$ $b=c=1$, $a=x$ for some $x\in [0,1]$, when $\textrm{LHS} = \dfrac {x+2}{x+1} + \sqrt[n]{x}$. But $2\sqrt{x}+1 \leq x+2$, since equivalent to $(\sqrt{x}-1)^2 \geq 0$, and $3\sqrt[3]{x} \leq 2\sqrt{x}+1$, since equivalent to $(\sqrt[6]{x}-1)^2(2\sqrt[6]{x} + 1) \geq 0$, thus the only one worth to be checked is $\dfrac {x+2}{x+1} + \sqrt[n]{x} \leq \dfrac {5}{2}$, equivalent to $\dfrac {1}{x+1} + \sqrt[n]{x} \leq \dfrac {3}{2}$. This expression on the $\textrm{LHS}$ is clearly increasing in $n$. Now, for $n=3$, denoting $t=\sqrt[3]{x}$, it is equivalent to $\dfrac {1}{t^3+1} + t \leq \dfrac {3}{2}$, or $2t^4 - 3t^3 + 2t - 1 \leq 0$, or again $(t-1)(2t^3-t^2-t+1)\leq 0$. But $0\leq t\leq 1$, and $2t^3-t^2-t+1 = 2t(t-1/2)^2 + (t-3/4)^2 + 7/16 > 0$, so the inequality holds true. However, for $n=4$, denoting $t=\sqrt[4]{x}$, it would be equivalent to $\dfrac {1}{t^4+1} + t \leq \dfrac {3}{2}$, or $2t^5 - 3t^4 + 2t - 1 \leq 0$, or again $(t-1)^2(2t^3+t^2-1)\leq 0$, clearly false for $t<1$ but $t$ close enough to $1$. Therefore the largest such $n$ is $\boxed{n=3}$. A nasty enough problem that it was a good thing it was not selected for the competition. Can you please give some another nice solution.

What with (a, b, c)=(0.99, 1, 1)?

mavropnevma wrote: A nasty enough problem that it was a good thing it was not selected for the competition. HoneySingh wrote: Can you please give some another nice solution. From the definition of a nasty problem there's no entry for your query. #1500

The problem does have a "somewhat" nice solution: (P.S. Taken from the brilliant site of Andrei Eckstein : "Pregatire Matematica Olimpiade Juniori" (pregatirematematicaolimpiadejuniori.wordpress.com)

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