نامساوی

$\left\{\begin{array}{l}Ali>Reza\\ Reza>Hasan\end{array}\right.〉Ali>Reza>Hasan$

$\frac{-1}{5}<\frac{1}{5}\Rightarrow -5<5$

نامساوی فوق صحیح است.

نمی‌توانیم طرفین وسطین کنیم چون نمی‌دانیم سمت چپ نامساوی عددی مثبت است یا منفی، بایستی تعیین علامت شود:

$\frac{1}{x-3}<\frac{1}{5}\Rightarrow \frac{1}{x-3}-\frac{1}{5}<0\Rightarrow \frac{5-\left(x-3\right)}{5\left(x-3\right)}<0\Rightarrow \frac{8-x}{5\left(x-3\right)}<0$

$D=\left(-\infty ,3\right)\cup \left(8,+\infty \right)$

توان $b$ زوج است، بنابراین مخرج همواره مثبت می‌باشد. 

$b^2>0\Rightarrow \left\{\begin{array}{l}b>0\\ b<0\end{array}\right.$

برای برقراری نابرابری بالا باید $a.c$ منفی باشد و این در صورتی اتفاقی می‌افتد که علامت $a$ و $c$ مختلف العلامه باشند:

$a.c<0\Rightarrow \left\{\begin{array}{l}a>0,c<0\\ a<0,c>0\end{array}\right.$

برای برقراری این نابرابری، علامت $b$ می‌تواند هم مثبت باشد و هم منفی باشد و علامت $a$ و $c$ مختلف العلامه باشند.

$\frac{a}{bc}>0\Rightarrow \left\{\begin{array}{l}a>0,bc>0\Rightarrow \left\{\begin{array}{l}b>0,c>0\\ b<0,c<0\end{array}\right.\\ a<0,bc<0\Rightarrow \left\{\begin{array}{l}b>0,c<0\\ b<0,c>0\end{array}\right.\end{array}\right.$

برای برقراری این نابرابری، باید $a$ و $b.c$ هم علامت باشند.

یادآوری)

$a^m<b^m\Rightarrow \left|a\right|<\left|b\right|;m=2k$

طبق قانون فوق، اگر $a^2>b^2$ باشد، نمی‌توان نتیجه گرفت $a>b$ است، اما می‌توان نتیجه گرفت که $\left|a\right|>\left|b\right|$ است.

$\begin{array}{l}\sqrt{a^2+b^2}\le \left|a\right|+\left|b\right|\\ \\ \Rightarrow {\left(\sqrt{a^2+b^2}\right)}^2\le {\left(\left|a\right|+\left|b\right|\right)}^2\\ \\ \Rightarrow \left(a^2+b^2\right)\le a^2+2\left|a\right|\left|b\right|+b^2\\ \\ \Rightarrow 2\left|a\right|\left|b\right|\ge 0\end{array}$

نامساوی $2\left|a\right|\left|b\right|\ge 0$ همواره برقرار است پس نامساوی مورد نظر همواره برقرار است.

$\begin{array}{l}a<b;k>0\\ \\ \Rightarrow ak<bk\\ \\ \Rightarrow ak+ab<bk+ab\\ \end{array}$

$\begin{array}{l}\Rightarrow a(k+b)<b(k+a)\\ \\ \Rightarrow \frac{a}{b}<\frac{a+k}{b+k}\end{array}$

$\begin{array}{l}{\left(a-b\right)}^2\ge 0\\ \\ \Rightarrow (a^2-2ab+b^2)\ge 0\\ \\ \Rightarrow a^2-ab+b^2\ge ab;a+b>0\\ \end{array}$

$\begin{array}{l}\Rightarrow (a+b)(a^2-ab+b^2)\ge (a+b)ab\end{array}$

$\begin{array}{l}\\ \Rightarrow a^3+b^3\ge ab(a+b)\\ \\ \Rightarrow 3(a^3+b^3)\ge 3ab(a+b)\\ \\ \Rightarrow 3a^3+3b^3\ge 3a^{2}b+3a{b}^2\end{array}$

$\Rightarrow 3a^3+3b^3+\left(a^3+b^3\right)\ge 3a^{2}b+3a{b}^2+\left(a^3+b^3\right)$

$\begin{array}{c}\\ \begin{array}{l}\Rightarrow 4a^3+4b^3\ge {\left(a+b\right)}^3\\ \\ \Rightarrow \frac{4a^3+4b^3}{8}\ge \frac{{\left(a+b\right)}^3}{8}\\ \\ \Rightarrow \frac{4(a^3+b^3)}{8}\ge \frac{{\left(a+b\right)}^3}{2^3}\\ \\ \Rightarrow \frac{a^3+b^3}{2}\ge {\left(\frac{a+b}{2}\right)}^3\end{array}\end{array}$

$\begin{array}{l}a,b>0\Rightarrow 1+ab>0\\ \\ \\ \frac{a+b}{1+ab}<1\\ \\ \Rightarrow (1+ab)\frac{a+b}{1+ab}<(1+ab)\times 1\\ \\ \Rightarrow a+b<1+ab\\ \\ \Rightarrow a+b-(1+ab)<0\\ \end{array}$

$\begin{array}{l}\Rightarrow a+b-1-ab<0\\ \\ \Rightarrow (a-ab)-(1-b)<0\\ \\ \Rightarrow a(1-b)-(1-b)<0\\ \\ \Rightarrow (1-b)(a-1)<0\end{array}$

$\left\{\begin{array}{l}0<b<1\Rightarrow 1-b>0\\ 0<a<1\Rightarrow a-1<0\end{array}\right.〉\left(1-b\right)\left(a-1\right)<0$

نامساوی $\left(1-b\right)\left(a-1\right)<0$ همواره برقرار است پس نامساوی مورد نظر همواره برقرار است. 

$\begin{array}{l}(\sqrt{a}+\sqrt{b})>\sqrt{a+b}\\ \\ \Rightarrow {\left(\sqrt{a}+\sqrt{b}\right)}^2>{\left(\sqrt{a+b}\right)}^2\\ \end{array}$

$\begin{array}{l}\Rightarrow (a+2\sqrt{a}\sqrt{b}+b)>a+b\\ \\ \Rightarrow 2\sqrt{a}\sqrt{b}>0\\ \\ \Rightarrow 2\sqrt{ab}>0\end{array}$

نامساوی $2\sqrt{ab}>0$ همواره برقرار است پس نامساوی مورد نظر همواره برقرار است. 

$\begin{array}{l}\Rightarrow \sqrt{a}+\sqrt{a+2}<\sqrt{a+1}+\sqrt{a+1}\end{array}$

$\begin{array}{l}\\ \Rightarrow \sqrt{a+2}-\sqrt{a+1}<\sqrt{a+1}-\sqrt{a}\end{array}$

$\begin{array}{c}\\ \begin{array}{l}\Rightarrow (\sqrt{a+2}-\sqrt{a+1})\cdot \frac{\left(\sqrt{a+2}+\sqrt{a+1}\right)}{\left(\sqrt{a+2}+\sqrt{a+1}\right)}<(\sqrt{a+1}-\sqrt{a}).\frac{\sqrt{a+1}+\sqrt{a}}{\sqrt{a+1}+\sqrt{a}}\end{array}\end{array}$

$\begin{array}{c}\\ \begin{array}{l}\Rightarrow \frac{(a+2)-(a+1)}{\sqrt{a+2}+\sqrt{a+1}}<\frac{(a+1)-\left(a\right)}{\sqrt{a+1}+\sqrt{a}}\end{array}\end{array}$

$\begin{array}{l}\\ \Rightarrow \frac{1}{\sqrt{a+2}+\sqrt{a+1}}<\frac{1}{\sqrt{a+1}+\sqrt{a}}\end{array}$

$\begin{array}{l}\\ \Rightarrow \sqrt{a+2}+\sqrt{a+1}>\sqrt{a+1}+\sqrt{a}\end{array}$

$\begin{array}{l}\\ \Rightarrow \sqrt{a+2}>\sqrt{a}\end{array}$

نامساوی $\sqrt{a+2}>\sqrt{a}$ همواره برقرار است پس نامساوی مورد نظر همواره برقرار است. 

$\begin{array}{l}\sqrt{n-k}+\sqrt{n+k}<\sqrt{n-m}+\sqrt{n+m}\end{array}$

$\begin{array}{l}\Rightarrow \sqrt{n+k}-\sqrt{n-m}<\sqrt{n+m}-\sqrt{n-k}\end{array}$

$\begin{array}{l}\Rightarrow \left(\sqrt{n+k}-\sqrt{n-m}\right).\frac{\left(\sqrt{n+k}+\sqrt{n-m}\right)}{\left(\sqrt{n+k}+\sqrt{n-m}\right)}<\left(\sqrt{n+m}-\sqrt{n-k}\right).\frac{(\sqrt{n+m}+\sqrt{n-k})}{(\sqrt{n+m}+\sqrt{n-k})}\end{array}$

$\begin{array}{l}\Rightarrow \frac{(n+k)-(n-m)}{(\sqrt{n+k}+\sqrt{n-m})}<\frac{(n+m)-(n-k)}{(\sqrt{n+m}+\sqrt{n-k})}\end{array}$

$\Rightarrow \frac{k+m}{(\sqrt{n+k}+\sqrt{n-m})}<\frac{m+k}{(\sqrt{n+m}+\sqrt{n-k})}$

$\begin{array}{l}\Rightarrow \frac{(\sqrt{n+k}+\sqrt{n-m})}{k+m}>\frac{(\sqrt{n+m}+\sqrt{n-k})}{m+k}\end{array}$

$\Rightarrow \sqrt{n+k}+\sqrt{n-m}>\sqrt{n+m}+\sqrt{n-k}$

$\left\{\begin{array}{l}k>m\Rightarrow \sqrt{n+k}>\sqrt{n+m}\\ k>m\Rightarrow \sqrt{n-m}>\sqrt{n-k}\end{array}〉\right.\sqrt{n+k}+\sqrt{n-m}>\sqrt{n+m}+\sqrt{n-k}$

نامساوی اخير همواره برقرار است پس نامساوی فوق هم همواره برقرار است.

$\begin{array}{l}\frac{1}{4}=\frac{1}{2\times 2}<\frac{1}{1\times 2}\\ \\ \frac{1}{9}=\frac{1}{3\times 3}<\frac{1}{2\times 3}\\ ⋮\\ \frac{1}{n^2}=\frac{1}{n\times n}<\frac{1}{(n-1)n}\end{array}$

طرفين نامساوی های فوق را با هم جمع می‌كنيم:

$\begin{array}{l}\frac{1}{4}+\frac{1}{9}+\cdot \cdot \cdot +\frac{1}{n^2}<\frac{1}{1\times 2}+\frac{1}{2\times 3}+\cdot \cdot \cdot +\frac{1}{(n-1)n}\end{array}$

$\begin{array}{l}\Rightarrow \frac{1}{4}+\frac{1}{9}+\cdot \cdot \cdot +\frac{1}{n^2}<\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\cdot \cdot \cdot +\frac{n-(n-1)}{(n-1)n}\end{array}$

$\begin{array}{l}\Rightarrow \frac{1}{4}+\frac{1}{9}+\cdot \cdot \cdot +\frac{1}{n^2}<\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdot \cdot \cdot +\left(\frac{1}{n-1}-\frac{1}{n}\right)\end{array}$

$\begin{array}{l}\Rightarrow \frac{1}{4}+\frac{1}{9}+\cdot \cdot \cdot +\frac{1}{n^2}<1-\frac{1}{n}<1\end{array}$

$\Rightarrow \frac{1}{4}+\frac{1}{9}+\cdot \cdot \cdot +\frac{1}{n^2}<1$

$2(\sqrt{n+1}-\sqrt{n})=2(\sqrt{n+1}-\sqrt{n}).\frac{(\sqrt{n+1}+\sqrt{n})}{(\sqrt{n+1}+\sqrt{n})}=\frac{2\left[(n+1)-n\right]}{(\sqrt{n+1}+\sqrt{n})}=\frac{2}{(\sqrt{n+1}+\sqrt{n})}$

$\begin{array}{l}\frac{2}{\sqrt{n+1}+\sqrt{n}}<\frac{2}{\sqrt{n}+\sqrt{n}}=\frac{2}{2\sqrt{n}}=\frac{1}{\sqrt{n}}\end{array}$

$\begin{array}{l}\Rightarrow \frac{2}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{\sqrt{n}}\end{array}$

$\begin{array}{l}\Rightarrow \frac{2}{\left(\sqrt{n+1}+\sqrt{n}\right)}\cdot \frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}-\sqrt{n}\right)}<\frac{1}{\sqrt{n}}\end{array}$

$\begin{array}{l}\Rightarrow \frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{(n+1)-n}<\frac{1}{\sqrt{n}}\end{array}$

$\begin{array}{l}\Rightarrow 2\left(\sqrt{n+1}-\sqrt{n}\right)<\frac{1}{\sqrt{n}}\\ \\ \Rightarrow \frac{1}{\sqrt{n}}>2(\sqrt{n+1}-\sqrt{n})\end{array}$

$\begin{array}{l}ifn=1\Rightarrow \frac{1}{\sqrt{1}}>2(\sqrt{1+1}-\sqrt{1})\Rightarrow 1>2(\sqrt{2}-1)\end{array}$

$\begin{array}{l}ifn=2\Rightarrow \frac{1}{\sqrt{2}}>2(\sqrt{2+1}-\sqrt{2})\Rightarrow \frac{1}{\sqrt{2}}>2(\sqrt{3}-\sqrt{2})\end{array}$

$\begin{array}{l}ifn=3\Rightarrow \frac{1}{\sqrt{3}}>2(\sqrt{3+1}-\sqrt{3})\Rightarrow \frac{1}{\sqrt{3}}>2(\sqrt{4}-\sqrt{3})\end{array}$

$\begin{array}{l}⋮\\ ifn=n\Rightarrow \frac{1}{\sqrt{n}}>2(\sqrt{n+1}-\sqrt{n})\end{array}$

طرفين نامساوی های فوق را با هم جمع می‌كنيم:

$\begin{array}{l}1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdot \cdot \cdot +\frac{1}{\sqrt{n}}>2(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+2(\sqrt{4}-\sqrt{3})+\cdot \cdot \cdot +2(\sqrt{n+1}-\sqrt{n})\end{array}$

$\begin{array}{l}\Rightarrow 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdot \cdot \cdot +\frac{1}{\sqrt{n}}>\left(2\sqrt{2}-2+2\sqrt{3}-2\sqrt{2}+2\sqrt{4}-2\sqrt{3}+\cdot \cdot \cdot +2\sqrt{n+1}-2\sqrt{n}\right)\end{array}$

$\begin{array}{l}\Rightarrow 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdot \cdot \cdot +\frac{1}{\sqrt{n}}>2\sqrt{n+1}-2\end{array}$

$\Rightarrow 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\mathrm{...}+\frac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-1\right)$

$\begin{array}{l}ifn>1\Rightarrow \frac{1}{n+1}>\frac{1}{2n}\\ \\ ifn>2\Rightarrow \frac{1}{n+2}>\frac{1}{2n}\\ ⋮\\ \frac{1}{n+n}=\frac{1}{2n}\end{array}$

طرفين نامساوی های فوق را جمع می‌كنيم:

$\begin{array}{l}\frac{1}{n+1}+\frac{1}{n+2}+\cdot \cdot \cdot +\frac{1}{n+n}>\frac{1}{2n}+\frac{1}{2n}+\cdot \cdot \cdot +\frac{1}{2n}=\frac{n}{2n}=\frac{1}{2}\end{array}$

$\begin{array}{l}\Rightarrow \frac{1}{n+1}+\frac{1}{n+2}+\cdot \cdot \cdot +\frac{1}{n+n}>\frac{1}{2}\end{array}$

$\Rightarrow \frac{1}{n+1}+\frac{1}{n+2}+\cdot \cdot \cdot +\frac{1}{2n}>\frac{1}{2}$

$\begin{array}{l}\frac{1}{n+1}+\frac{1}{n+2}+\cdot \cdot \cdot +\frac{1}{3n+1}>(4n+2)\left\{\frac{1}{2{(2n+1)}^2}\right\}\end{array}$

$\begin{array}{l}\Rightarrow \frac{1}{n+1}+\frac{1}{n+2}+\cdot \cdot \cdot +\frac{1}{n+n}+\frac{1}{2n+1}+\frac{1}{2n+2}+\mathrm{...}+\frac{1}{2n+n}+\frac{1}{3n+1}>\frac{1}{2n+1}\end{array}$

$\Rightarrow \frac{1}{n+1}+\frac{1}{n+2}+\cdot \cdot \cdot +\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}+\mathrm{...}+\frac{1}{3n}+\frac{1}{3n+1}>\frac{1}{2n+1}$

$\begin{array}{l}\Rightarrow \left(\frac{1}{n+1}+\frac{1}{3n+1}\right)+\left(\frac{1}{n+2}+\frac{1}{3n}\right)+\cdot \cdot \cdot +\left(\frac{1}{2n}+\frac{1}{2n+2}\right)+\frac{1}{2n+1}>\frac{1}{2n+1}\end{array}$

$\begin{array}{l}\Rightarrow \frac{4n+2}{(n+1)(3n+1)}+\frac{4n+2}{(n+2)3n}+\cdot \cdot \cdot +\frac{4n+2}{2n(2n+2)}+\frac{1\times 2(2n+1)}{2(2n+1)(2n+1)}>\frac{1}{2n+1}\end{array}$

$\Rightarrow \frac{4n+2}{(n+1)(3n+1)}+\frac{4n+2}{(n+2)3n}+\cdot \cdot \cdot +\frac{4n+2}{2n(2n+2)}+\frac{4n+2}{2{(2n+1)}^2}>\frac{1}{2n+1}$

$\begin{array}{l}\Rightarrow (4n+2)\left[\frac{1}{(n+1)(3n+1)}+\frac{1}{(n+2)3n}+\cdot \cdot \cdot +\frac{1}{2n\left(2n+2\right)}+\frac{1}{2{(2n+1)}^2}\right]>\frac{1}{2n+1}\end{array}$

$\Rightarrow \frac{1}{n+1}+\frac{1}{n+2}+\cdot \cdot \cdot +\frac{1}{3n+1}>\frac{1}{2n+1}$

$\begin{array}{l}\frac{1}{2}<\frac{2}{3}\\ \\ \frac{3}{4}<\frac{4}{5}\\ \\ \frac{5}{6}<\frac{6}{7}\\ ⋮\\ \frac{99}{100}<\frac{100}{101}\end{array}$

طرفين نامساوی های فوق را در هم ضرب كنيم:

$\frac{1}{2}\times \frac{3}{4}\times \frac{5}{6}\times \cdot \cdot \cdot \times \frac{99}{100}<\frac{2}{3}\times \frac{4}{5}\times \frac{6}{7}\times \cdot \cdot \cdot \times \frac{100}{101}$

$\begin{array}{l}ifx=\frac{1}{2}\times \frac{3}{4}\times \frac{5}{6}\times \cdot \cdot \cdot \times \frac{99}{100}\end{array}$

$\begin{array}{l}if\frac{1}{x}=\frac{2}{1}\times \frac{4}{3}\times \frac{6}{5}\times \cdot \cdot \cdot \times \frac{100}{99}\end{array}$

$\Rightarrow \frac{1}{x}=\frac{2}{3}\times \frac{4}{5}\times \cdot \cdot \cdot \times \frac{98}{99}\times \frac{100}{1}$

$\Rightarrow \frac{1}{101x}=\frac{2}{3}\times \frac{4}{5}\times \cdot \cdot \cdot \times \frac{100}{101}$

$\begin{array}{l}x<\frac{1}{101x}\\ \\ \Rightarrow x^2<\frac{1}{101}\\ \\ \Rightarrow x<\frac{1}{\sqrt{101}}<\frac{1}{10}\end{array}$

$\Rightarrow x<\frac{1}{10}\Rightarrow \frac{1}{2}\times \frac{3}{4}\times \frac{5}{6}\times \cdot \cdot \cdot \times \frac{99}{100}<\frac{1}{10}$

$\begin{array}{l}x-\frac{ax+by}{a+b}=\frac{x(a+b)-(ax+by)}{a+b}\\ \\ \Rightarrow x-\frac{ax+by}{a+b}=\frac{ax+bx-ax-by}{a+b}\end{array}$

$\begin{array}{l}\Rightarrow x-\frac{ax+by}{a+b}=\frac{bx-by}{a+b}\\ \\ \Rightarrow x-\frac{ax+by}{a+b}=\frac{b(x-y)}{a+b}\\ \\ \Rightarrow x-\frac{ax+by}{a+b}=\frac{b(x-y)}{a+b};\left(1\right)\end{array}$

$\begin{array}{l}x>y\end{array}$

$\begin{array}{l}\Rightarrow x-y>0;a,b>0,a+b>0\end{array}$

$\begin{array}{l}\Rightarrow \frac{b(x-y)}{a+b}>0;\left(1\right)\\ \\ \Rightarrow x-\frac{ax+by}{a+b}>0\\ \\ \Rightarrow x>\frac{ax+by}{a+b};\left({\rm I}\right)\end{array}$

$\begin{array}{l}y-\frac{ax+by}{a+b}=\frac{y(a+b)-(ax+by)}{a+b}\end{array}$

$\Rightarrow y-\frac{ax+by}{a+b}=\frac{ay+by-ax-by}{a+b}$

$\begin{array}{l}\Rightarrow y-\frac{ax+by}{a+b}=\frac{ay-ax}{a+b}\\ \\ \Rightarrow y-\frac{ax+by}{a+b}=\frac{a(y-x)}{a+b}\\ \\ \Rightarrow y-\frac{ax+by}{a+b}=\frac{a(y-x)}{a+b};\left(2\right)\end{array}$

$\begin{array}{l}x>y\end{array}$

$\begin{array}{l}\Rightarrow y-x<0;a,b>0,a+b>0\end{array}$

$\begin{array}{l}\Rightarrow \frac{a(y-x)}{a+b}<0;\left(2\right)\\ \\ \Rightarrow y-\frac{ax+by}{a+b}<0\\ \\ \Rightarrow y<\frac{ax+by}{a+b};\left({\rm I}{\rm I}\right)\end{array}$

$\left({\rm I}\right) , \left({\rm I}{\rm I}\right)\Rightarrow \left\{\begin{array}{l}x>\frac{ax+by}{a+b}\\ y<\frac{ax+by}{a+b}\end{array}\right.〉y<\frac{ax+by}{a+b}<x$

$\begin{array}{l}0\le x\le 1\Rightarrow 1+2x^3>0\\ \\ \\ \frac{1}{1+x^2+x^4}\le \frac{1}{1+2x^3}\\ \\ \Rightarrow (1+x^2+x^4)\ge (1+2x^3)\\ \end{array}$

$\begin{array}{l}\Rightarrow x^4-2x^3+x^2\ge 0\\ \\ \Rightarrow x^2(x^2-2x+1)\ge 0\\ \\ \Rightarrow x^2{\left(x-1\right)}^2;\left({\rm I}\right)\end{array}$

نامساوی $\left({\rm I}\right)$ همواره برقرار است.

$\begin{array}{l}\frac{1}{1+x^2+x^4}\ge \frac{1}{1+2x^2}\\ \\ \Rightarrow \left(1+x^2+x^4\right)\le \left(1+2x^2\right)\end{array}$

$\begin{array}{l}\Rightarrow x^4-x^2\le 0\\ \\ \Rightarrow x^2\left(x^2-1\right)\le 0\\ \\ \Rightarrow x^2\left(x-1\right)\left(x+1\right)\le 0;\left({\rm I}{\rm I}\right)\end{array}$

با توجه به $0\le x\le 1$ نامساوی فوق همواره برقرار است، پس صورت مسئله برقرار است.

$\begin{array}{l}ifx=\sqrt[3]{3}+\sqrt[3]{9}\\ \\ \Rightarrow x^3={\left(\sqrt[3]{3}+\sqrt[3]{9}\right)}^3\end{array}$

$\begin{array}{l}\Rightarrow x^3={\left(\sqrt[3]{3}\right)}^3+3{\left(\sqrt[3]{3}\right)}^2\left(\sqrt[3]{9}\right)+3\left(\sqrt[3]{3}\right){\left(\sqrt[3]{9}\right)}^2+{\left(\sqrt[3]{9}\right)}^3\end{array}$

$\begin{array}{l}\Rightarrow x^3=3+3\sqrt[3]{3^2\times 9}+3\sqrt[3]{3\times 9^2}+9\end{array}$

$\begin{array}{l}\Rightarrow x^3=12+3\sqrt[3]{3^3\times 3}+3\sqrt[3]{3^3\times 3^2}\end{array}$

$\begin{array}{l}\Rightarrow x^3=12+\left(3\times 3\sqrt[3]{3}\right)+\left(3\times 3\sqrt[3]{9}\right)\end{array}$

$\begin{array}{l}\Rightarrow x^3=12+9\left(\sqrt[3]{3}+\sqrt[3]{9}\right)\\ \\ \Rightarrow x^3=12+9x\\ \\ \Rightarrow x^3-9x-12=0\end{array}$

$x=\sqrt[3]{3}+\sqrt[3]{9}$ ريشه حقيقی تابع زیر است:

$f\left(x\right)=x^3-9x-12=x(x^2-9)-12$

اگر $x>3$ تابع $f\left(x\right)$ صعودی است، از طرف ديگر طبق قضیه بولتزوانو داریم:

$\left\{\begin{array}{l}f\left(\sqrt[3]{43}\right)<0\\ f\left(\sqrt[3]{49}\right)>0\end{array}\right.〉\sqrt[3]{43}<\sqrt[3]{3}+\sqrt[3]{9}<\sqrt[3]{49}$

$\begin{array}{l}\frac{x^3+x}{x-1}\ge 2\left(2+\sqrt{3}\right)\\ \\ \Rightarrow \frac{x^3+x}{x-1}-\left(4+2\sqrt{3}\right)\ge 0\\ \\ \Rightarrow \frac{\left(x^3+x\right)-\left(x-1\right)\left(4+2\sqrt{3}\right)}{x-1}\ge 0\end{array}$

$\begin{array}{l}\Rightarrow \frac{x^3+x-\left(4x+2\sqrt{3}x-4-2\sqrt{3}\right)}{x-1}\ge 0\end{array}$

$\Rightarrow \frac{x^3+x-4x-2\sqrt{3}x+4+2\sqrt{3}}{x-1}\ge 0$

چون $x>1$ است عبارت صورت و مخرج هر دو مثبت است پس نامساوی اخير درست است، بنابراين نامساوی صورت مسئله برقرار است.

$\begin{array}{l}{n}^4>a^4\Rightarrow n>a\Rightarrow n\ge a+1\Rightarrow n^4\ge {\left(a+1\right)}^4\end{array}$

$\begin{array}{l}{m}^4>b^4\Rightarrow m>b\Rightarrow m\ge b+1\Rightarrow m^4\ge {\left(b+1\right)}^4\end{array}$

$k^4>c^4\Rightarrow k>a\Rightarrow k\ge c+1\Rightarrow k^4\ge {\left(c+1\right)}^4$

$\begin{array}{l}\left\{\begin{array}{l}{a}^4+14ab+1=n^4\ge {\left(a+1\right)}^4\\ b^4+14bc+1=m^4\ge {\left(b+1\right)}^4\\ c^4+14ac+1=k^4\ge {\left(c+1\right)}^4\end{array}\right.\end{array}$

$\begin{array}{l}{a}^4+14ab+1=n^4\ge {\left(a+1\right)}^4\end{array}$

$\begin{array}{l}\Rightarrow a^4+14ab+1\ge {\left(a+1\right)}^4\end{array}$

$\begin{array}{l}\Rightarrow a^4+14ab+1\ge a^4+4a^3+6a^2+4a+1\end{array}$

$\begin{array}{l}\Rightarrow 14ab\ge 4a^3+6a^2+4a\end{array}$

$\begin{array}{l}\Rightarrow 14ab\ge 2a\left(2a^2+3a+2\right)\end{array}$

$\Rightarrow 7b\ge 2a^2+3a+2$

به‌همین ترتیب داریم:

$\left\{\begin{array}{l}7b\ge 2a^2+3a+2\\ 7c\ge 2b^2+3b+2\\ 7a\ge 2c^2+3c+2\end{array}\right.$

طرفین نامساوی های فوق را باهم جمع می‌کنیم: 

$\begin{array}{l}7a+7b+7c\ge 2a^2+2b^2+2c^2+3a+3b+3c+6\end{array}$

$\begin{array}{l}\Rightarrow 2a^2+2b^2+2c^2-4a-4b-4c+6\le 0\end{array}$