已知正实数 $a,b,c$ 满足 $\min\{ab,bc,ca\}\geqslant 1$．证明：$$\sqrt[3]{\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)} \leqslant\left(\frac{a+b+c}{3}\right)^{2}+1$$
【难度】
【出处】
2016IMO Short List
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【答案】
先证明一个引理．
引理：若任意正实数 $x,y$ 满足 $xy\geqslant 1$，则$$\left(x^{2}+1\right)\left(y^{2}+1\right) \leqslant\left(\left(\frac{x+y}{2}\right)^{2}+1\right)^{2}$$由 $xy\geqslant 1$，知$$\left(\frac{x+y}{2}\right)^{2}-1 \geqslant x y-1 \geqslant 0$$则$$\begin{aligned}\left(x^{2}+1\right)\left(y^{2}+1\right)&=(x y-1)^{2}+(x+y)^{2} \\ & \leqslant\left(\left(\frac{x+y}{2}\right)^{2}-1\right)^{2}+(x+y)^{2} \\ &=\left(\left(\frac{x+y}{2}\right)^{2}+1\right)^{2}\end{aligned}$$引理得证．
不妨设 $a\geqslant b\geqslant c$．这表明，$a\geqslant 1$．设 $d=\frac{a+b+c}{3}$．则$$a d=\frac{a(a+b+c)}{3} \geqslant \frac{1+1+1}{3}=1$$对于数对 $(a,d),(b,c)$，由引理得$$\left(a^{2}+1\right)\left(d^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \leqslant\left(\left(\frac{a+d}{2}\right)^{2}+1\right)^{2}\left(\left(\frac{b+c}{2}\right)^{2}+1\right)^{2}$$注意到，$\frac{a+d}{2} \cdot \frac{b+c}{2} \geqslant \sqrt{a d} \sqrt{b c} \geqslant 1$．对于数对 $\left(\frac{a+d}{2},\frac{b+c}{2}\right)$，由引理得$$\begin{aligned}&\left(a^{2}+1\right)\left(d^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)\leqslant\left(\left(\frac{a+b+c+d}{4}\right)^{2}+1\right)^{4}=\left(d^{2}+1\right)^{4}\\ &\Rightarrow\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \leqslant\left(d^{2}+1\right)^{3}\end{aligned}$$上式两边同时开三次根号即得原不等式成立
引理：若任意正实数 $x,y$ 满足 $xy\geqslant 1$，则$$\left(x^{2}+1\right)\left(y^{2}+1\right) \leqslant\left(\left(\frac{x+y}{2}\right)^{2}+1\right)^{2}$$由 $xy\geqslant 1$，知$$\left(\frac{x+y}{2}\right)^{2}-1 \geqslant x y-1 \geqslant 0$$则$$\begin{aligned}\left(x^{2}+1\right)\left(y^{2}+1\right)&=(x y-1)^{2}+(x+y)^{2} \\ & \leqslant\left(\left(\frac{x+y}{2}\right)^{2}-1\right)^{2}+(x+y)^{2} \\ &=\left(\left(\frac{x+y}{2}\right)^{2}+1\right)^{2}\end{aligned}$$引理得证．
不妨设 $a\geqslant b\geqslant c$．这表明，$a\geqslant 1$．设 $d=\frac{a+b+c}{3}$．则$$a d=\frac{a(a+b+c)}{3} \geqslant \frac{1+1+1}{3}=1$$对于数对 $(a,d),(b,c)$，由引理得$$\left(a^{2}+1\right)\left(d^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \leqslant\left(\left(\frac{a+d}{2}\right)^{2}+1\right)^{2}\left(\left(\frac{b+c}{2}\right)^{2}+1\right)^{2}$$注意到，$\frac{a+d}{2} \cdot \frac{b+c}{2} \geqslant \sqrt{a d} \sqrt{b c} \geqslant 1$．对于数对 $\left(\frac{a+d}{2},\frac{b+c}{2}\right)$，由引理得$$\begin{aligned}&\left(a^{2}+1\right)\left(d^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)\leqslant\left(\left(\frac{a+b+c+d}{4}\right)^{2}+1\right)^{4}=\left(d^{2}+1\right)^{4}\\ &\Rightarrow\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \leqslant\left(d^{2}+1\right)^{3}\end{aligned}$$上式两边同时开三次根号即得原不等式成立
【解析】
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