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![](https://rs.olm.vn/images/avt/0.png?1311)
Tham khảo:
https://hoc24.vn/cau-hoi/cho-a-b-c-la-do-dai-ba-canh-cua-mot-tam-giac-va-thoa-man-he-thuc-a-b-c-1-cmr-a2-b2-c2-12.139261258302
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì a,b,c là 3 cạnh tam giác nên \(a+b>c\Leftrightarrow ac+bc>c^2\)
CMTT: \(ab+bc>b^2;ab+ac>a^2\)
Cộng vế theo vế \(\Leftrightarrow a^2+b^2+c^2< ab+bc+ca+ab+bc+ca\)
\(\Leftrightarrow a^2+b^2+c^2< 2ab+2bc+2ca\\ \Leftrightarrow a^2+b^2+c^2-2ab-2bc-2ca< 0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Từ gt suy ra a < b + c nên 2a < a + b + c = 2
\(\Rightarrow a< 1\).
Chứng minh tương tự: \(b< 1;c< 1\).
Do đó \(\left(a-1\right)\left(b-1\right)\left(c-1\right)< 0\Leftrightarrow abc< ab+bc+ca-1\) (Do a + b + c = 2)
\(\Rightarrow a^2+b^2+c^2+2abc< a^2+b^2+c^2+2\left(ab+bc+ca-1\right)=\left(a+b+c\right)^2-2=2\) (đpcm).
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT tam giác, ta có:
\(\hept{\begin{cases}a< b+c\\b< c+a\\c< a+b\end{cases}\Rightarrow\hept{\begin{cases}2a< a+b+c\\2b< a+b+c\\2c< a+b+c\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}2a< 6\\2b< 6\\2c< 6\end{cases}\Rightarrow\hept{\begin{cases}a< 3\\b< 3\\c< 3\end{cases}\Rightarrow}}\hept{\begin{cases}3-a>0\\3-b>0\\3-c>0\end{cases}}\)
Áp dụng BĐT Cauchy cho bộ ba số thực không âm, ta có:
\(\left(3-a\right)\left(3-b\right)\left(3-c\right)\le\left(\frac{3-a+3-b+3-c}{3}\right)^3=1\)
\(\Leftrightarrow27-9\left(a+b+c\right)+3\left(ab+bc+ca\right)-abc\le1\)
\(\Leftrightarrow abc\ge27-9.6+3\left(ab+bc+ca\right)-1\)
\(\Leftrightarrow2abc\ge-56+6\left(ab+bc+ca\right)\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)+2abc\ge3\left(a^2+b^2+c^2\right)+3.2\left(ab+bc+ca\right)-56\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)+2abc\ge3\left(a+b+c\right)^2-56\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)+2abc\ge3.36-56=\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)+2abc\ge52\)
Dấu \("="\) xảy ra khi \(a=b=c=2\)
Vậy \(3\left(a^2+b^2+c^2\right)+2abc\ge52\)
Theo bất đẳng thức tam giác: \(\left\{{}\begin{matrix}a+b>c\Leftrightarrow a+b+c>2c\Leftrightarrow2c< 2\Leftrightarrow c< 1\\b+c>a\Leftrightarrow a+b+c>2a\Leftrightarrow2a< 2\Leftrightarrow a< 1\\a+c>b\Leftrightarrow a+b+c>2b\Leftrightarrow2b< 2\Leftrightarrow b< 1\end{matrix}\right.\Leftrightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)>0\)\(\Rightarrow\left(1-b-a+ab\right)\left(1-c\right)>0\) \(\Rightarrow1-c-b+bc-a+ac+ab-abc>0\) \(\Rightarrow1+bc+ac+ab>2+abc\Leftrightarrow bc+ac+ab>1+abc\) \(\Rightarrow2ab+2bc+2ac>2+2abc\Leftrightarrow\left(a+b+c\right)^2>2+2abc+a^2+b^2+c^2\) \(\Rightarrow a^2+b^2+c^2+2abc+2< 4\Leftrightarrow a^2+b^2+c^2+2abc< 2\)(đpcm)