neravenstvo

[N.Stavrev]

Да се докаже, че за произволни положителни [tex]a,b,c[/tex] с произведение 1 е изпълнено
[tex]a+b+c+3\le \frac{1}{2 }\sum_{cyc}^{ }\sqrt{(a+b)((a+b)(c+1)+4)}[/tex]

[rashi101]

[tex]\frac{1}{2 }\sum_{cyc}^{ }\sqrt{(a+b)((a+b)(c+1)+4)}=\\
\frac{1}{2 }\sum_{cyc}^{ }\sqrt{(a+b)((a+b)(\frac{1}{ab } +1)+4)}=\\
\frac{1}{2 }\sum_{cyc}^{ }\sqrt{(a+b)(\frac{1}{a } + \frac{1}{b }+a+b+4)}=\\
\frac{1}{2}\sum_{cyc}^{}\sqrt{(a^2+b^2+2ab+4a+4b+2+\frac{a}{b }+\frac{b}{ a})}\ge\\
\frac{1}{2 }\sum_{cyc}^{ }\sqrt{(a^2+b^2+2ab+4a+4b+2+2)}=\\
\frac{1}{2 }\sum_{cyc}^{ }\sqrt{(a+b+2)^2}=a+b+c+3[/tex]

Равенство при [tex]a=b=c[/tex]