1．已知$x_1,x_2,\cdots ,x_{n+1}$是$n+1$个正实数，证明：$$\dfrac{1}{x_1}+\dfrac{x_1}{x_2}+\dfrac{x_1x_2}{x_3}+\cdots +\dfrac{x_1x_2\cdots x_n}{x_{n+1}}\geqslant 4\left(1-x_1x_2\cdots x_{n+1}\right).$$
2．已知$a,b,c\in\mathcal R$，且$a+b+c=1$，$a^2+b^2+c^2=1$，求证：$a^5+b^5+c^5\leqslant 1$．
3．设$x,y,z\in\mathcal R$，$k>0$，求证：$4\left(x^2+k\right)\left(y^2+k\right)\left(z^2+k\right)\geqslant 3k^2\left(x+y+z\right)^2$．
4．在$\triangle ABC$中，$A,B,C$所对的边分别为$a,b,c$，若$a+b+c=1$，求证：$a^2+b^2+c^2+4abc<\dfrac 12$．
5．已知$x,y,z>0$，且$x+y+z=1$，求证：$\sqrt{1+\dfrac{yz}{x}}+\sqrt{1+\dfrac{zx}{y}}+\sqrt{1+\dfrac{xy}z}\geqslant 2\sqrt 3$．
6．设$x,y,z>0$，求证：$\dfrac 18(x+y)(y+z)(z+x)\geqslant \dfrac 13(x+y+z)(xyz)^{\frac 23}$．
7．设$a,b,c>0$，且$ab+bc+ca=1$，求证：$\sum\limits_{{cyc}}\displaystyle \sqrt[3]{\dfrac 1a+6b}\leqslant \dfrac{1}{abc}$．
参考答案
1．注意到对任意正实数$x$，均有$$\dfrac{1}{x}\geqslant 4(1-x),$$于是\[\begin{split} LHS&\geqslant 4\left(1-x_1\right)+4x_1\left(1-x_2\right)+4x_1x_2\left(1-x_3\right)+\cdots +4x_1x_2\cdots x_n\left(1-x_{n+1}\right)\\ &=RHS,\end{split} \]于是原不等式得证．
2．根据题意，$a,b,c\in [-1,1]$且有$$ab+bc+ca=\dfrac{(a+b+c)^2-\left(a^2+b^2+c^2\right)}{2}=0,$$于是\[\begin{split} 1-\left(a^5+b^5+c^5\right)&=(a+b+c)^5-\left(a^5+b^5+c^5\right)\\&=5(a+b)(b+c)(c+a)\left(a^2+b^2+c^2+ab+bc+ca\right)\\ &=5(1-a)(1-b)(1-c) \\ &\geqslant 0,\end{split} \]因此命题得证．
3．换元，令$a=\dfrac{x\sqrt 2}{\sqrt k}$，$b=\dfrac{y\sqrt 2}{\sqrt k}$，$c=\dfrac{z\sqrt 2}{\sqrt k}$，则原不等式等价于$$\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)\geqslant 3(a+b+c)^2.$$由均值不等式和柯西不等式有\[\begin{split} LHS&=\left(a^2b^2+1+2a^2+2b^2+3\right)\left(c^2+2\right)\\&\geqslant \left(2ab+2a^2+2b^2+3\right)\left(c^2+2\right)\\&\geqslant\left[\dfrac 32(a+b)^2+3\right]\left(c^2+2\right)\\&=\dfrac 32\left[(a+b)^2+2\right]\left(2+c^2\right)\\&\geqslant \dfrac 32\left[(a+b)\cdot \sqrt 2+\sqrt 2\cdot c\right]^2\\&=RHS,\end{split} \]因此原不等式得证．
4．原命题等价于$$(a+b+c)^3>2(a+b+c)\left(a^2+b^2+c^2\right)+8abc,$$即$$\sum_{cyc}{a^2(b+c-a)}>2abc.$$令$x+y=a,y+z=b,x+z=c$，则上式转化为$$(x+y)^2z+(y+z)^2x+(z+x)^2y>(x+y)(y+z)(z+x),$$展开整理得到$4xyz>0$，故原命题得证．
5．原不等式等价于$$3+\sum_{cyc}\dfrac{yz}x+2\sum_{cyc}\sqrt{1+\dfrac{yz}x+\dfrac{zx}y+z^2}\geqslant 12,$$而$$\sum_{cyc}\dfrac{yz}x=\dfrac 12\sum_{cyc}\left(\dfrac{yz}x+\dfrac{zx}y\right)\geqslant \dfrac 12\sum_{cyc}(2z)=1,$$又$$\sum_{cyc}\sqrt{1+\dfrac{yz}x+\dfrac{zx}y+z^2}\geqslant \sum_{cyc}\sqrt{1+2z+z^2}=\sum_{cyc}(1+z)=4,$$因此原命题得证．
6．不妨设$x+y+z=1$，则由$$(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)-xyz,$$可得原不等式等价于$$xy+yz+zx-xyz\geqslant \dfrac 83\left(xyz\right)^{\frac 23}.$$由排序不等式知$$\left(xy+yz+zx\right)^2\geqslant 3\sum_{cyc}(xy^2z)=3xyz\sum_{cyc}x=3xyz,$$于是只需要证明$$\sqrt{3xyz}-xyz\geqslant \dfrac 83\left(xyz\right)^{\frac 23},$$也即$$\dfrac 83\left(xyz\right)^{\frac 16}+\left(xyz\right)^{\frac 12}\leqslant \sqrt 3.$$而$$\left(xyz\right)^{\frac 13}\leqslant \dfrac{x+y+z}3=\dfrac 13,$$于是上述不等式成立，原不等式得证．
7．由幂平均不等式，有$$LHS\leqslant 3\left[\dfrac{\sum_{cyc}\left(\dfrac 1a+6b\right)}{3}\right]^{\frac 13},$$于是只需要证明$$9\sum_{cyc}\left(\dfrac 1a+6b\right)\leqslant \left(\dfrac{1}{abc}\right)^3.$$由切比雪夫不等式知$$\left(\dfrac{1}{abc}\right)^2=\left(\dfrac 1a+\dfrac 1b+\dfrac 1c\right)^2\geqslant 3\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right),$$于是$$\dfrac{1}{abc}\geqslant 3(a+b+c),$$于是$$9\sum_{cyc}\left(\dfrac 1a+6b\right)=9\left(\dfrac 1{abc}+6(a+b+c)\right)\leqslant \dfrac{27}{abc},$$而$$\dfrac{1}{abc}=\dfrac 1a+\dfrac 1b+\dfrac 1c\geqslant 3\sqrt[3]{\dfrac{1}{abc}},$$于是$\left(\dfrac{1}{abc}\right)^2\geqslant 27$，因此原命题得证．